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Vol. XVII No. 3 March 2015 Corporate Office: Plot 99, Sector 44 Institutional Area, Gurgaon -122 003 (HR), Tel : 0124-4951200 e-mail : [email protected] website : www.mtg.in

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The content for PMT Biology is very vast and does not allow students to engage in inquiry and develop meaningful knowledge. An essential topic for PMT is presented here to enable students grasp the topic, analyse the type of questions appearing in PMTs, and score HIGH.

TransporT in planTs -ii TRANSLOCATION OF WATER ANd mINERALS (ASCENT OF SAp) The upward movement of sap (water with dissolved ingredients) from the root towards the top of the plant is known as ascent of sap. Roots are organs concerned with the absorption of water. If the plant is to survive and grow, this water absorbed by roots must reach the top of plants to replace the water lost in transpiration and to be used in photosynthesis as a raw material. The upward movement of water occurs through the tracheary elements (tracheids and vessels) of xylem. Sap is lifted from near the root tip to the shoot tip against the force of gravity, sometimes to great heights. Some trees are much taller, often attaining a height upto 110-130 m, e.g., Sequoia sempervirens (height 110 m), Picea sitchensis (height 95-100 m), Pseudotsuga menziesii (height 100 m), Eucalyptus (height above 130 m), etc. The force required to move water to such heights are substantive. The rate of translocation is 25–75 cm/min (15–45 m/hr). Several theories have been put forward to explain the mechanism of ascent of sap. The four main theories are: vital force theory, relay pump theory, root pressure theory, capillary force theory and transpiration pull theory.

Vital force theory A common vital force theory about the ascent of sap was put forward by Sir J.C. Bose in 1923. It is called pulsation theory and it believes that the innermost cortical cells of the root absorb water from the outer side and pump the same into xylem channels. 8

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Relay pump theory Relay pump theory was put forward by Godlewski (1884). According to this, upward conduction of water is due to the pumping activity of xylem parenchyma cells and the cells of medullary rays.

Root pressure theory Root pressure theory was put forward by Priestley (1916). Root pressure refers to positive hydrostatic pressure which sometimes develops in the xylem sap of roots as a result of metabolic activities of roots. It is a manifestation of active water absorption. Root pressure is observed in certain seasons which favour optimum metabolic activity and reduce transpiration. Root pressure is maximum during rainy season (in tropical regions) and spring season (in temperate habitats). It is retarded or absent under conditions of starvation, low temperature, drought and reduced availability of oxygen. Root pressures are usually not more than + 1 to + 2 bars. Therefore, it could account for the ascent of sap only to a height of about 20 m. Hence, this theory can account for the ascent of sap only in the herbaceous plants. The magnitude of pressure developed is too small to push the water to the apical regions, in the tall trees. Besides, root pressure is not found in all plants. No or little root pressure is found in gymnosperms which have some of the tallest trees in the world. In rapidly transpiring plants, no root pressure is found, instead a negative pressure is found under such conditions.

Capillary force theory Capillary force theory was given by Boehm (1809). According to this theory, water rises up in narrow tubes of xylem vessels by surface tension or adhesion and cohesion. The upward movement of water will continue till the forces of adhesion and cohesion are balanced by the downward pull of gravity. The value of capillarity is very small which can raise water to a height of about 1 metre in vessels of normal diameter (0.03 mm). Therefore, if operational, it will be useful to only small sized plants and cannot operate in plants having tracheids due to the presence of end walls.

Transpiration pull or cohesion-tension theory The most widely accepted theory for ascent of sap is transpiration pull theory or cohesion-tension theory. This theory was proposed by Dixon and Jolly in 1894. This theory is based on the following points: Transpirational pull exerted on the water column. – – Cohesive and adhesive properties of water molecules so as to form an unbroken continuous column of water in the xylem.

Transpiration pull Water is lost from mesophyll cells to the intercellular spaces of leaves as a result of transpiration. The water vapours move out of the plant through stomata. As a result of loss of water from mesophyll cells, the diffusion pressure deficit (DPD) increases. 10

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With the increase of diffusion pressure deficit, these cells absorb water from adjoining cells, ultimately the water is absorbed from xylem elements of vascular bundles of leaf. Since the xylem elements are filled with continuous water column, a tension or pull called transpiration pull develops at the top of the column. This tension or pull is transmitted down from petiole to stem and finally to roots leading to upward movement of water.

Cohesion and adhesion of water in xylem Xylem tracheids and trachea are long tubular structures filled with water, extending from root to leaf. Thus, one end of xylem (continuous with one another) is in the root and other end is in the leaf. Water molecules remain attached to one another by a strong mutual force of attraction called cohesive force, which is due to the presence of hydrogen bonds amongst adjacent water molecules. Supplementing the cohesion between water molecules is adhesion between water molecules and the walls of tracheary elements of xylem. Thus, according to this theory, water ascends in the plant because of transpiration pull and this column of water remains continuous because of cohesive and adhesive forces of water molecules.

TRANSpIRATION The loss of water in the form of vapours from the living tissues of aerial parts of the plant is termed as transpiration. The loss of water due to transpiration is quite high, 2 litres per day in sunflower, 36-45 litres in apple and upto 1 tonne per day in elm tree. Approximately 98-99% of the water absorbed by a plant is lost in transpiration, 0.2% is used in photosynthesis while the remaining is retained in the plant during growth. Most of the transpiration occurs through foliar surface or surface of the leaves. It is known as foliar transpiration. Young stems, flowers, fruits, etc., also transpire. Transpiration from stems is called cauline transpiration. Different types of transpiration have been listed in the following flow chart.

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Stomatal apparatus The stomata are tiny pores present in the epidermal surface of leaves, young stems and in certain fruits (e.g., banana, Citrus, cucumber, etc.). This pore is surrounded by kidney shaped or bean shaped epidermal cells called guard cells. In monocots, the guard cells generally are dumb-bell shaped. These cells are living, having nucleus, chloroplasts and cytoplasm. The walls of guard cells are thickened on inner side. They have one or two pairs of wall extensions or ledges to prevent entry of water drops into stomata. The outer walls of guard cells are thin and more elastic. These guard cells are surrounded by some specialized epidermal cells called subsidiary cells or accessory cells. Stomata are meant for gaseous exchange during photosynthesis and respiration, as well as they are the main source of transpiration. When turgidity increases within the two guard cells flanking each stomatal aperture or pore, the thin outer walls bulge out and force the inner walls into a crescent shape. The opening of the stomata is also aided due to the orientation of the microfibrils in the cell walls of the guard cells. Cellulose microfibrils are oriented radially rather than longitudinally, making it easier for the stomata to open. When the guard cells lose turgor, due to water loss (or water stress) the elastic inner walls regain their original shape, the guard cells become flaccid and the stomata close.

Number and distribution of stomata on leaf On the basis of number and distribution of stomata, leaves are categorized into following types: – Apple type (mulberry type): Stomata are present only on the lower surface of leaf (hypostomatic leaf), e.g., apple. – Potato type : Stomata are present on both surfaces of leaf but more on the lower surface, e.g., potato, pea, tomato and many other dicot plants. Oat type: The number of stomata is equal on both surfaces of leaf (amphistomatic leaves) e.g., oat and – many other monocots. – Water lily type: The stomata are present only on the upper surface of leaf, e.g., water lily and most floating plants (Nelumbo, Nymphaea, Victoria, Eurale, etc.). Potamogeton type : Stomata are either absent or vestigeal (astomatic leaves), e.g., Potamogeton, Hydrilla, – Najas.

period of stomatal opening On the basis of periods of opening and closing, stomata are of following types: – Alfalfa type: Stomata remain open throughout the day, remain closed at night, e.g., beans, pea. – Barley type : Stomata open only for a few hours in day and remain closed in rest period as in cereals. – Potato type: Stomata remain open throughout the day and night closing only under the period of water stress, when rate of transpiration exceeds the rate of absorption. – –

Equisetum type: Stomata seldom close, usually remain open throughout, as in emergent hydrophytes. Succulent xerophytes like Opuntia or Bryophyllum type: Stomata remain closed in day hours to discourage transpiration but open at night for gaseous exchange and CO2 fixation. MT BIOLOGY

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mechanism of stomatal movement Stomata function as turgor operated valves because opening and closing of stomata is governed by change in O.P. or turgidity of guard cells. When guard cells are turgid, stomata open and when guard cells are flaccid, stomata close. Different theories about the mechanism of stomatal movements have been proposed : Hypothesis of guard cell photosynthesis (given by Schwendener, 1881). – Starch-sugar interconversion theory (given by Lloyd, 1908 and by Sayre 1926 and modified by Steward – 1964). Malate or K+ ion pump hypothesis (given by Fujino, 1959 and modified by Levitt, 1974) –

Hypothesis of guard cell photosynthesis According to this, the photosynthetic activity of the guard cells was believed to be responsible for the stomatal movement. During the day, glucose is photosynthesised and being osmotically active, lowers the water potential of the cell sap, thus there is an influx of water into the guard cells and the stomata open. At night, sugar is converted to starch and being osmotically inactive, decreases the osmotic concentration of guard cell. Thus, water moves out and guard cells become flaccid and stomata close. Biggest objection to it was that the photosynthetic activity of guard cell chloroplasts was found to be negligible.

Starch-sugar interconversion theory According to this theory, a change in pH effects the opening and closing of stomata. Starch sugar interconversion is brought about by a pH dependent enzyme starch phosphorylase. The change in the pH of the guard cell’s sap is due to the presence or absence of CO2 which is dependent on light availability. During day time, in the presence of light, CO2 is utilized for photosynthesis and hence does not accumulate in the guard cells and pH of the guard cells increases to 7. Hence, the enzyme favours the formation of glucose-1-phosphate from starch and inorganic phosphate.

Glucose being osmotically active, lowers the water potential of the cell sap, thus there is an influx of water into the guard cells and the stomata open. At night, in the absence of light, CO2 evolved in respiration accumulates in the cell sap and dissolves in water to form carbonic acid and pH of the cell sap decreases to 5 (acidic). At this pH, starch synthesis is favoured from glucose-1-phosphate.

Starch being osmotically inactive, decreases the osmotic concentration of guard cell. Thus, water moves out and guard cells become flaccid and stomata close. There are many objections to starch-sugar interconversion theory. First of all, it is a slow process and cannot account for rapid stomatal movements. Secondly, enzyme phosphorylase catalyzes the reaction only in forward direction. Moreover, sugar has never been found in free state in guard cells of open stomata and starch has never been reported from guard cells of onion. 14

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malate or K+ ion pump hypothesis This is the most accepted modern theory. It is also known as hormonal regulation theory (due to ABA secretion).

This theory may be represented as follows:

In light First of all, the starch is converted into organic acids particularly phosphoenol pyruvic acid (PEP). Phosphoenol pyruvic acid then combines with CO2 in the presence of PEP carboxylase enzyme to produce oxalo acetic acid (OAA) and then malic acid. The organic acids, viz., malic acid, dissociate into malate anion and H+ in the guard cells. H+ are transported to epidermal cells and K+ are taken into the guard cells in exchange of H+. The process is called ion exchange. K+ are balanced by organic anions (i.e., malate). Some Cl– ions are also taken in to neutralize a small percentage of K+. H+– K+ exchange is an active process which requires involvement of energy (ATP) supplied either by respiration or photophosphorylation. Increased concentration of K+ and malate ions in the vacuole of guard cells causes sufficient osmotic pressure to absorb water from surrounding cells. Increased turgor of guard cells due to entry of water causes stomatal pore to open.

In dark Higher concentration of CO2 in sub-stomatal cavity prevents proton gradient across protoplasmic membrane in guard cells (Sharpe and Zeiger, 1981). As a result, active transport of K+ into guard cells ceases. Cowan et al. (1982) proposed that closure mechanism involves participation of an inhibitor hormone-abscisic acid, which functions at lower pH. As soon as the pH of guard cells decreases, the abscisic acid inhibits K+ uptake by changing the diffusion and permeability of the guard cells. Malate ions present in the guard cell cytoplasm combine with H+ to produce MT BIOLOGY

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malic acid. Excess of malic acid inhibits its own synthesis by decreasing the activity of PEP carboxylase. These changes cause reversal of ion movement so that K+ is transported out of guard cells into surrounding epidermal cells. The osmotic concentration of guard cells is thus decreased, resulting in the movement of water from guard cells to surrounding cells. The guard cells become flaccid and the stomatal pore gets closed.

Factors affecting stomatal movement –

In majority of plants, the stomata open in light and close in darkness. Maximum opening occurs in red light and blue light. No opening occurs in green light, UV-light and infrared light.

–

– – – – – – –

Increase in temperature leads to opening of stomata upto certain extent. Very high temperature leads to closing of stomata even in day (mid-day closure). At 38° – 40°C, stomata can open in complete darkness, while at 0°C they remain closed even in continuous light. Normally high temperature above 30°C, reduces stomatal opening in many species. Water stress (or water deficit) brings about stomatal closure due to rise in DPD of epidermal cells. Mechanical shock causes closure of stomata. High CO2 concentration in the intercellular spaces of leaves causes closure of stomata even in day time. Low CO2 concentration usually induces opening of stomata. Rise in pH results in opening of stomata while fall in pH induces closing. Cytokinins are essential for opening of stomata while ABA takes part in stomatal closure. A number of minerals are essential for stomatal movements e.g., K, P, N, Mg, Ca etc. Oxygen is essential for stomatal opening.

Stomatal pore is measured with the help of instrument called porometer. Rate of transpiration is measured by potometer, e.g., Ganong’s potometer, Farmer’s potometer, Darwin’s potometer, etc. Principle of working of all potometers is that absorption is proportional to transpiration. Cobalt chloride paper method is used to compare rate of transpiration on two surfaces of leaf.

Factors affecting transpiration External factors – –

–

–

–

Light increases transpiration through opening of stomata and increased protoplasmic permeability. Increase in temperature brings about an increase in the rate of transpiration. Rate of transpiration is generally doubled with every 10°C rise in temperature. The rate of transpiration is inversely proportional to the relative humidity, i.e., the rate of transpiration is higher when the relative humidity is lower and vice versa. The movement of air increases the rate of transpiration by removing the saturated air around the leaves. Upto 20-30 km/hr, the rate of transpiration increases with the wind velocity. A wind velocity of 40-50 km/hr decreases transpiration by closing the stomata due to mechanical effect, drying and cooling of the transpiring organs. Reduced availability of soil water causes wilting or loss of turgidity resulting in drooping and rolling thus reduces transpirational rate.

Internal factors – – – –

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In general, higher the root/shoot ratio, more will be the rate of transpiration. It is due to the fact that an extensive root system is more efficient in water uptake from the soil. More is the leaf area, more will be the rate of transpiration. Cuticular transpiration decreases with the thickness of cuticle. The sunken (deep seated) stomata, present in xerophytes, are a device to reduce the rate of transpiration.

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– – –

Leaf modifications such as formation of prickles, leaf spines, phyllodes etc., help in reducing the rate of transpiration. Compact mesophyll reduces transpiration while a loose mesophyll increases transpiration. Presence of mucilage decreases the rate of transpiration by holding water. Wilting

Wilting is the loss of turgidity of leaves and other soft aerial parts of a plant causing their drooping, folding and rolling. The symptoms of wilting are not shown by thick-walled tissues. Therefore, they are less conspicuous in sclerophyllous plants. Wilting is of 3 types : incipient wilting, temporary or transient wilting and permanent wilting. In incipient wilting, there are no external symptoms of wilting but the mesophyll cells have lost sufficient water due to transpiration being higher than the availability of water. It occurs during mid day for a brief period in almost all plants even when sufficient water is present in the soil. Temporary or transient wilting is the temporary drooping down of leaves and young shoots due to loss of turgidity during noon. Lower leaves show wilting earlier than the upper ones. Temporary wilting is corrected only after the rate of transpiration decreases in the afternoon accompanied by replenishment of water around the root hairs. A permanent wilting is that state in the loss of turgidity of leaves when they do not regain their turgidity even on being placed in a saturated atmosphere. It occurs when the soil is unable to meet the requirement of plant for transpiration. After permanent wilting, the plant dies.

Transpiration as necessary evil Transpiration causes loss of huge amount of water absorbed by plants and leads to wilting and injury in plants. It also checks photosynthesis, reduces growth and if too severe, may cause death due to desiccation. Inspite of various detrimental effects, the plants cannot avoid transpiration due to their peculiar structure of leaves which is basically meant for gaseous exchange during respiration and photosynthesis and also for the required pull for ascent of sap. Therefore, transpiration is also regarded as “necessary evil” by Curtis (1926) or “unavoidable evil” by Steward (1959).

Antitranspirants Plant antitranspirants are chemical substances, whether naturally synthesised by plants themselves or synthetic materials applied to plants, which reduce the rate of transpiration effectively. An ideal antitranspirant is one which decreases the transpiration rate without affecting CO2 fixation in photosynthesis. There are two types of antitranspirants: metabolic inhibitors and surface films. Metabolic inhibitors reduce transpiration by reducing MT BIOLOGY

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the stomatal opening for a period of two or more weeks without influencing other metabolic processes. The most promising of these inhibitors is phenyl mercuric acetate. Another is abscisic acid (ABA). Film forming chemicals check transpiration by forming a thin film on the transpiring surface. They are sufficiently permeable to carbon dioxide and oxygen to allow photosynthesis and respiration but prevent movement of water vapours through them. The important chemicals of this group are silicon emulsions, colourless plastic resins and low viscosity waxes.

Transpiration and photosynthesis - A compromise An actively photosynthesizing plant has an insatiable need for water. Photosynthesis is limited by available water which can be swiftly depleted by transpiration. The humidity of rainforests is largely due to this vast cycling of water from root to leaf to atmosphere and back to the soil. The evolution of the C4 photosynthetic system is probably one of the strategies for maximizing the availability of CO2 while minimizing water loss. C4 plants are twice as efficient as C3 plants in terms of fixing carbon (making sugar). However, a C4 plant loses only half as much water as a C3 plant for the same amount of CO2 fixed.

GuTTATION Loss or excretion of water in the form of liquid droplets from the tips and margins of leaves is called guttation. It was first studied by Bergerstein in 1887. All plants do not show guttation. It is restricted to about 345 genera of herbaceous and some woody plants. Common examples are garden nasturtium, oat and other cereals, balsam, tomato, cucurbits etc. It occurs in warm, moist soils with humid environment or when warm days are followed by cool nights. In general, guttation occurs when transpiration rate is very low as compared to rate of water absorption. Due to this, root pressure is developed and water is pushed out through specialized pores at vein endings called hydathodes. So guttation is not due to activity of hydathodes but due to root pressure. Each hydathode consists of a group of loosely arranged colourless parenchymatous cells called epithem. It lies over the tip of a vascular strand and communicates with the outside through a permanent pore in epidermis called water pore or water stoma. The guttated liquid is never pure water. It contains 0.6-2.5 gm/litre of solutes– both organic (carbohydrates, organic 2– acids, amino acids, enzymes) and inorganic (Ca2+, Mg2+, K+, Na+, CO3 , SO42–, Cl–). These salts sometimes are redissolved back into leaves and cause ‘salt injury’. Gutttion takes place either at night or early in the morning. Dry soils, poorly aerated soils, heavily salted or mineral deficient soils and the atmospheric conditions promoting transpiration inhibit guttation. Table :

Differences between transpiration and guttation Transpiration

Guttation

1.

Loss of water is in vapour form.

Loss of water is in liquid form.

2.

It occurs during day time.

It occurs during night or early morning.

3.

Loss of pure water.

Loss of impure water.

4.

Through stomata or epidermis or cuticle or lenticels.

Through hydathodes.

5.

Controlled phenomenon.

Uncontrolled phenomenon.

6.

Associated with regulation of temperature.

No such role has been given to guttation.

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BLEEdING It is the exudation of sap or watery solution from the cut or injured parts of the plant, e.g., Agave, Acer, Vitis, etc. It occurs due to root pressure, phloem pressure, local pressure in xylem (stem pressure) and latex or resin.

upTAKE ANd TRANSpORT OF mINERAL NuTRIENTS uptake of mineral ions Unlike water, all minerals cannot be passively absorbed by the roots. Two factors account for this: (i) minerals are present in the soil as charged particles (ions) which cannot move across cell membranes and, (ii) the concentration of minerals in the soil is usually lower than the concentration of minerals in the root. Therefore, most minerals must enter the root by active absorption into the cytoplasm of epidermal cells. This needs energy in the form of ATP. The active uptake of ions is partly responsible for the water potential gradient in roots, and therefore for the uptake of water by osmosis. Some ions also move into the epidermal cells passively. Ions are absorbed from the soil by both passive and active transport. Specific proteins in the membranes of root hair cells actively pump ions from the soil into the cytoplasm of the epidermal cells. Like all cells, the endodermal cells have many transport proteins embedded in their plasma membrane; they let some solutes cross the membrane, but not others. Transport proteins of endodermal cells are control points, where a plant adjusts the quantity and types of solutes that reach the xylem. Note that the root endodermis because of the layer of suberin has the ability to actively transport ions in one direction only.

Translocation of mineral ions After the ions have reached xylem of the roots through active or passive uptake, or a combination of the two, their further transport up the stem to all parts of the plants is through the transpiration stream. Though it is generally considered that xylem transports inorganic nutrients while phloem transports organic nutrients, the same is not exactly true. In xylem sap, nitrogen travels as inorganic ions, as well as organic form of amino acids and related compounds. Small amounts of P and S are passed in xylem as organic compounds. There is also exchange of materials between xylem and phloem. Therefore, mineral elements pass up xylem in both inorganic and organic form. They reach the area of their sink, namely young leaves, developing flowers, fruits and seeds, apical and lateral meristems and individual cells for storage. Minerals are unloaded at fine vein endings through diffusion. They are picked up by cells through active uptake. There is remobilization of minerals from older senescing parts. Nickel has a prominent role in this activity. The senescing leaves send out many minerals like nitrogen, sulphur, phosphorous and potassium. Elements incorporated in structural components are, however, not remobilized, e.g., calcium. The remobilized minerals become available to young growing leaves and other sinks.

phLOEm TRANSpORT: TRANSLOCATION OF ORGANIC SOLuTES Formation of organic food or carbohydrate occurs by green parts of the plant (leaves) as a result of photosynthesis. Non-green parts depend upon green parts for organic food and thus food is transported from green parts to non-green parts of the plant. This movement of organic food or solute in soluble form, from one organ to another organ is called translocation of solutes. For example: from leaves to stem and roots for consumption. Translocation of solutes takes place always from higher concentration (source or supply end) of its soluble form to lower concentration of its soluble form (sink or consumption end). Translocation of solutes mostly occurs in form of sucrose. MT BIOLOGY

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Since the source-sink relationship is variable, the direction of movement in the phloem can be upwards or downwards, i.e., bi-directional. This contrasts with that of the xylem where the movement is always unidirectional, i.e., upwards. Hence, unlike one-way flow of water in transpiration, food in phloem sap can be transported in any required direction so long as there is a source of sugar and a sink able to use, store or remove the sugar.

direction of translocation of solutes Solutes are translocated in the plants in downward, upward or radial direction. Path of upward and downward translocation is phloem, while path of radial translocation is medullary rays. Downward translocation : It is of most important type, i.e., from leaves to stem and roots. Upward translocation : From leaves to developing flowers, buds, fruits and also during germination of seeds and tubers, etc. Radial translocation : From pith to cortex and epidermis.

Evidences in favour of downward conduction of solutes by phloem Downward conduction of solutes takes place through phloem is evident from the following points. Suitability : Phloem has long tubes (most suitable for long distance transport) placed one above the other. Thus, phloem seems to be the only suitable tissue for downward conduction of solutes. Chemical analysis of phloem sap : Chemical analysis of sieve tube sap proves that concentrated solution of sucrose is translocated from the place of synthesis to other parts of the plant body. Glucose and fructose are sometimes found in traces only. The amount of sucrose is more in phloem sap during the day and less in night. Isotopic studies : If leaf of a potted plant is illuminated in the presence of radioactive C14O2, it forms radioactive products of photosynthesis which are then transported to stem. It was detected by autoradiographic studies that these substances are translocated through phloem particularly sieve tubes. Radioactivity is found below and above the nodes of the leaf to which radioactive carbon was provided. Ringing or girdling experiment : In this experiment, all the tissues of the stem outside the vascular cambium are removed in a ring. The upper part of the plant is attached to lower part only by external xylem cylinder and pith. Accumulation of food occurs in the form of swelling just above the ring, which suggests that in absence of phloem, downward translocation of food is stopped. In a girdled plant, roots die first and ultimately shoot dies. This is because the upper part of stem gets ample amount of water supply (as transport of water and minerals occurs through xylem). But as the roots die due to starvation, the upper part of stem also dies in the course of time.

mechanism of translocation of solutes Several theories have been put forward to explain the mechanism of translocation of organic nutrients through phloem, but Munch mass flow hypothesis or pressure flow hypothesis is most accepted one. It was given by Munch (1930) and supported by Crafts. According to this theory, translocation of solutes takes place in bulk, through phloem along a gradient of turgor pressure from higher concentration of its soluble form (source) to lower concentration of its soluble form (sink). Because in leaves (source), osmotic concentration remains always high (due to photosynthesis) and in roots (sink), osmotic concentration remains low, so mass flow of organic food continues from leaves to roots (i.e., along a gradient of turgor pressure). 20

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The organic substances present in mesophyll cells are passed into the sieve tubes through their companion cells by an active process. A high osomotic concentration, therefore, develops in the sieve tubes of the source. Thus, the sieve tubes absorb water from the surrounding xylem and develop a high turgor pressure. It causes the flow of organic solution towards the area of low turgor pressure. A low turgor pressure is maintained in the sink region by converting soluble organic substances into insoluble form. Objections to this theory are as follows: Vacuoles of the adjacent sieve tube cells are not – continuous. The cytoplasm present near the sieve plates exerts resistance to the mass flow. Catalado et al (1972) have observed the rate of – flow of water (72 cm/hr) and solutes (35 cm/hr) to be different in the same sieve tube. Phloem transport is not influenced by water deficit. – – The cells at the source end of mass flow should be turgid but they are often found to be flaccid in case of germinating tubers, corms, etc. Various factors affecting translocation of solutes have been summarized in the given flow chart.

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New MCQs 1. Ringing/girdling experiments demonstrate (a) phloem is responsible for translocation of food (b) xylem is responsible for ascent of sap (c) transpiration pull (d) both (a) and (b). 2. Which of the following theories gives the latest explanation for the opening and closure of stomata? (a) Starch-sugar conversion theory (b) Munch theory (c) Transpiration pull theory (d) Active K+ transport theory 3. Match column-I with column-II and select the correct option from the codes given below. Column-I Column-II (i) Rate of transpiration A. Dixon and Jolly B. Stomata closure (ii) Transpiration pull C. Potometer (iii) Only water available to plants (iv) ABA D. Capillary water (a) A-(i), B-(iv), C-(ii), D-(iii) (b) A-(ii), B-(iv), C-(i), D-(iii) (c) A-(ii), B-(i), C-(iii), D-(iv) (d) A-(iii), B-(ii), C-(i), D-(iv) 4. Which of the following statements is incorrect? (a) Guard cells are kidney shaped in dicots and dumb bell shaped in monocots. (b) Free-floating hydrophytes possess epistomatic leaves. (c) In xerophytes such as Opuntia, stomata are absent. (d) Amphistomatic leaves are generally present in monocots. 5. The transpiration-driven ascent of xylem sap depends mainly upon _____ property of water. (a) cohesion (b) adhesion (c) surface tension (d) all of these 6. Which of the following is correct regarding the translocation of substances in vascular tissues of plants? (a) Organic substances, e.g., sugars are transported upward in the xylem. (b) Organic substances move up and down in phloem. (c) Salts and other inorganic substances move downward only through the xylem. (d) Inorganic substances move upward only through the pholem. 22

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7. A girdled plant (upto bast) may survive for some time but it will eventually die, because (a) water will not move downwards (b) water will not move upwards (c) sugars and other organic materials will not move downwards (d) sugars and other organic materials will not move upwards. 8. Guttation is the release of liquid water from veins at the leaf margins. It is caused by (a) transpiration (b) high leaf pressure (c) high root pressure (d) clogged tracheids or vessel elements. 9. Wilting of a plant occurs when (a) xylem is blocked (b) phloem is blocked (c) epidermis and few roots are removed (d) pith is removed. 10. Translocation of minerals takes place through (a) phloem (b) xylem (c) cambium (d) wood parenchyma. 11. The continuity of water column in xylem is maintained due to (a) adhesive property of water (b) cohesive property of water (c) presence of air bubbles (d) both (a) and (b). 12. Stomata of a plant open when guard cells show (a) influx of hydrogen ions (b) efflux of potassium ions (c) influx of potassium ions (d) all of these. 13. Potometer works on the principle of (a) osmotic pressure (b) amount of water absorbed equals the amount transpired (c) root pressure (d) capillarity. 14. The rate of transpiration of a plant would gradually increase if (a) the relative humidity increases (b) the relative humidity decreases (c) the relative humidity remains unchanged (d) the water potential gradient remains unchanged.

15. The process of guttation takes place when (a) the root pressure is high and the rate of transpiration is low (b) the root pressure is low and the rate of transpiration is high (c) the root pressure equals the rate of transpiration (d) the root pressure as well as rate of transpiration are high. 16. Select the mismatched pair. (a) Relay pump theory – Godlewski (b) Pulsation theory – Sir J. C. Bose (c) Transpiration pull theory – Dixon and Jolly (d) Capillary force theory – Munch 17. Unidirectional flow of water, minerals, some organic nitrogen and hormones occurs through (a) xylem (b) phloem (c) root (d) vascular tissue. 18. Match column-I with column-II and select the correct option from the codes given below. Column-I Column-II A. Vein ending (i) Transpiration B. Necessary evil (ii) Osmosis C. Semipermeable (iii) Transpiration pull membrane D. Cohesion (iv) Guttation E. Stomata closure (v) ABA (a) A-(iv), B-(i), C-(iii), D-(ii), E-(v) (b) A-(iv), B-(i), C-(ii), D-(iii), E-(v) (c) A-(iii), B-(v), C-(i), D-(ii), E-(iv) (d) A-(i), B-(ii), C-(iii), D-(iv), E-(v) 19. Loss of water in the form of vapours from the living tissues of aerial plant parts is called as (a) transpiration (b) guttation (c) bleeding (d) precipitation. 20. Rate of transpiration increases with an increase in (a) temperature (b) light (c) wind velocity (d) all of these.

Exam Section 1. The shade of a tree is cooler than the shade of a roof due to (a) transpiration (b) guttation (c) photosynthesis (d) green leaves. (AIIMS 1995) 2. Transpiration can be influenced by interfering with (a) osmotic pressure (b) guard cell (c) atmospheric temperature (d) epidermis of leaf. (AFMC 1995)

3. Sir J.C. Bose proposed which of the following theories to explain the process of ascent of sap ? (a) Pulsation theory (b) Relay pump theory (c) Transpiration pull theory (d) Capillary force theory (Karnataka1996) 4. The plants face wilting due to use of excessive fertilizers, because of (a) exosmosis (b) imbibition (c) endosmosis (d) all of these. (UP-CPMT 1998) 5. Opening and closing of stomata is due to the (a) hormonal change in guard cells (b) change in turgor pressure of guard cells (c) gaseous exchange (d) respiration. (CBSE-PMT 2002) 6. “Transpiration is a necessary evil”, this statement belongs to (a) Burgerstein (b) Reschke (d) Willmer. (c) Curtis (Manipal 2002) 7. Opening or closing of stomata are under the control of (a) K+ (b) P (c) Mg (d) none of these. (DPMT 2002) 8. Water lost by transpiration is (a) rich in dissolved minerals (b) rich in solutes (c) pure water (d) rich in dissolved salts (Karnataka 2002) 9. Guttation is loss of water (a) in the form of vapour (b) in the form of liquid droplets (c) from root (d) through stomata. (Manipal 2004) 10. Transpiration is very important for plants because it helps in (a) the absorption of water from soil (b) the cooling of leaves at high temperature (c) the movement of water and minerals absorbed by roots to various parts of the plant (d) all of the above. (AMU 2004) MT BIOLOGY

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11. The following figure shows the stomatal apparatus. Identify the parts labelled as A, B, C and D and choose the correct answer from the following.

(a) A = Guard cells, B = Stoma, C = Chloroplasts, D = Subsidiary cells (b) A = Subsidiary cells, B = Chloroplasts, C = Stoma, D = Guard cells (c) A = Guard cells, B = Chloroplasts, C = Stoma, D = Subsidiary cells (d) A = Subsidiary cells, B = Stoma, C = Chloroplasts, D = Guard cells (Karnataka 2005) 12. In which of the following plants, there will be no transpiration? (a) Aquatic, submerged plants (b) Plants living in deserts (c) Aquatic plants with floating leaves (d) Plants growing in hilly regions (Karnataka 2005) 13. Upward movement of water in plants is called (a) sucking (b) ascent of sap (c) translocation (d) none of these. (Odisha 2006) 14. Sunken stomata are found in (a) xerophytes (b) hydrophytes (c) mesophytes (d) all of these. (Odisha 2006) 15. The rupture and fractionation do not usually occur in the water column in vessel/tracheids during the ascent of sap because of (a) weak gravitational pull (b) transpiration pull (c) lignified thick walls (d) cohesion and adhesion. (CBSE-PMT 2008) 16. Which of the following chemical serves as an antitranspirant in plants? (a) Cobalt chloride (b) Dimethyl mercury (c) Potassium iodide (d) Phenyl mercuric acetate (BHU 2008) 24

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17. Which of the following does not cause opening of stomata? (a) Light (b) Increased pH (c) Reduced pH (d) Low CO2 concentration (DPMT 2008) 18. Stomatal opening or closing is assisted by (a) change in the turgidity of guard cells (b) the inner walls of each guard cells which are thick and elastic (c) cellulose microfibrils of guard cells, oriented radially (d) all of the above. (AMU 2010) 19. Which of the following is not a purpose of transpiration? (a) Transports minerals from the soil to all parts of the plant (b) Helps in translocation of sugars from source to sink (c) Maintains shape and structure of the plants (d) Cools leaf surfaces (Kerala 2010) 20. In land plants, the guard cells differ from other epidermal cells in having (a) cytoskeleton (b) mitochondria (c) endoplasmic reticulum (d) chloroplasts. (AIPMT Prelims 2011) 21. Accumulation of which one of the following acids results into closure of stomata ? (a) Malic acid (b) Aspartic acid (c) Phosphoenol pyruvic acid (d) Oxaloacetic acid (AMU 2011) 22. Radial conduction of water takes place by (a) vessels (b) vessels and trachieds (c) phloem (d) ray parenchyma cells. (AMU 2012) 23. Force generated by transpiration can create pressure sufficient to lift water even upto the height of (a) 130 feet (b) 130 metre (c) 230 feet (d) 230 metre. (AMU 2012) 24. Stomata open at night in (a) hydrophytes (b) succulents (c) mesophytes (d) halophytes. (Odisha 2012)

25. Which of the following statements is not true for stomatal apparatus? (a) Guard cells invariably possess chloroplasts and mitochondria. (b) Guard cells are always surrounded by subsidiary cells. (c) Stomata are involved in gaseous exchange. (d) Inner wall of guard cells are thick. (NEET - Karnataka 2013)

Assertion & Reason The following questions consist of two statements each : assertion (A) and reason (R). To answer these questions, mark the correct alternative as directed below : (a) If both A and R are true and R is the correct explanation of A. (b) If both A and R are true but R is not the correct explanation of A. (c) If A is true but R is false. (d) If both A and R are false. 1. Assertion : Exudation of sap from cut or injured parts of plants is called bleeding. Reason : Bleeding occurs due to root pressure, phloem pressure and local pressure in xylem. 2. Assertion : Loss of water in the form of liquid droplets from leaves of plants is called transpiration. Reason : Transpiration takes place through special structures called hydathodes. 3. Assertion : Xylem transport is unidirectional. Reason : Phloem transport is bi-directional. 4. Assertion : Transpiration is an unavoidable evil. Reason : Plants cannot regulate transpiration. 5. Assertion : Guttation drops are restricted to tips or margins of leaves while dew drops are found all over the plant. Reason : Both dew drops and guttated water are pure water.

Short Answer Type Questions 1. Read the following statements and fill in the blanks with correct words/terms. (i) Rate of transpiration is _____ when relative humidity is _____. (ii) _____ and _____ are antitranspirants which reduce transpiration by reducing the stomatal opening for a period of few weeks, without influencing other metabolic activities.

(iii) The movement of ions from soil to interior of root requires _____. (iv) Mineral elements pass up xylem in both _____ and _____ form. (v) In_____ there is no external symptom, but mesophyll cells lose sufficient water due to transpiration being higher than the availability of water. 2. How do surface films check transpiration? 3. What adaptations are found in xerophytic plants to reduce the rate of transpiration?

Answer Key New MCQs 1. 6. 11. 16.

(d) (b) (d) (d)

2. 7. 12. 17.

(d) (c) (c) (a)

3. 8. 13. 18.

(b) (c) (b) (b)

4. 9. 14. 19.

(c) (a) (b) (a)

5. 10. 15. 20.

(d) (b) (a) (d)

(a) (c) (b) (d) (b)

4. 9. 14. 19. 24.

(a) (b) (a) (b) (b)

5. 10. 15. 20. 25.

(b) (d) (d) (d) (b)

Exam Section 1. 6. 11. 16. 21.

(a) (c) (c) (d) (a)

2. 7. 12. 17. 22.

(c) (a) (a) (c) (d)

3. 8. 13. 18. 23.

Assertion & Reason 1. (a)

2. (d)

3. (b)

4. (c)

5. (c)

Short Answer Type Questions 1. (i) higher/lower, lower/higher (ii) Phenyl mercuric acetate, ABA (iii) active transport (iv) inorganic, organic (v) incipient wilting 2. Film forming chemicals check transpiration by forming a thin film on the transpiring surface. They are sufficiently permeable to carbon dioxide and oxygen and allow photosynthesis and respiration but prevent movement of water vapours through them. 3. The sunken (deep seated) stomata and leaf modifications such as formation of prickles, leaf spines, phyllodes etc., are some adaptations in xerophytes which help in reducing the rate of transpiration. nn MT BIOLOGY

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1.

The given graph shows the concentration of urine produced in three different mammals, A, B and C.

Three types of kidney tubules, X, Y and Z, are also given below.

i.

What is the reason for the production of different concentrations of urine in these mammals?

-----------------------------------------------------------------------------------. ii. Refer the graph and match the kidney tubules (X, Y and Z) with the mammals (A, B and C) to which they belong. ----------------------------------------------------------------------------------.

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Length of loop of Henle may vary in different kidney tubules and is related to urine concentration.

2.

i.

ii.

3.

One way to express some aspects of the age distribution characteristics of a population is through a survivorship curve. The given graph shows the percentage of maximum life span and survival per thousand of three organisms i.e., humans, Hydra and oyster.

The type I, type II and type III survivorship curves in the given graph, belong to which organisms respectively? ----------------------------------------------------------------------------------. Explain why, these three organisms have different survivorship curves.

Survivorship is defined as percentage of an original population that survives to a given age.

----------------------------------------------------------------------------------. Given below are the steps involved in two processes P and Q of an assay.

i.

Explain the principle involved in the assay and identify the processes P and Q.

ii.

----------------------------------------------------------------------------------------------------------------------------------------. Explain the significance of the two processes of the assay. ----------------------------------------------------------------------------------------------------------------------------------------. MT BIOLOGY

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4.

There are several species of North American hares. The given figure shows the correlation between external ear size in different species of hares and latitude.

i.

Explain why arctic hare has smallest external ear size?

ii.

----------------------------------------------------------------------------------. Identify the rule, associated with given figure?

Arctic hare is adapted largely to cold habitats.

----------------------------------------------------------------------------------.

5.

Given figure shows the protein structure of the enzyme cellulase, which is found in grass eating animals so that they can digest the cellulose found in grass.

i.

Name structure A and describe how it is formed.

ii.

----------------------------------------------------------------------------------------------------------------------------------------. State the differences between the structures of cellulase and its substrate. ----------------------------------------------------------------------------------------------------------------------------------------.

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sOLuTIOns 1. (i) Mammals are capable of forming urine, that is more concentrated than blood plasma, thus maximising the amount of water retained. The nephron of mammals is able to do this because it possesses a loop of Henle. The loop of Henle has a key role in water conservation, as good amount of water is reabsorbed from the glomerular filtrate over here. Based on lengths of the loop of Henle, there are two types of nephrons: (i) cortical nephrons: they have short loops of Henle which only just extend into the medulla and function most importantly when the water supply is normal to excessive, (ii) juxtamedullary nephrons : they have long loops of Henle that extend deep into the medulla and function most importantly when water is in short supply. The lengths of loops of Henle and of the collecting ducts, and the general thickness of the medulla region of the kidneys, increase progressively in animals best adapted to the drier habitats. (ii) Kidney tubule X belongs to mammal B, kidney tubule Y belongs to mammal A and kidney tubule Z belongs to mammal C. According to the graph, mammal C is producing most concentrated urine which indicates that it is adapted to arid habitats and possesses kidney with juxtamedullary nephrons. Mammal A is producing least concentrated urine which implies that it thrives well in aquatic habitat so it possesses kidney with cortical nephrons. Mammal B is producing intermediate concentration of urine which indicates that it is well adapted to terrestrial habitats, with adequate water supply hence, possesses kidney tubules intermediate between the two. 2. (i) Type I survivorship curve belongs to humans, type II belongs to Hydra and type III belongs to oyster. (ii) Type I surviorship curves are characterised by high age-specific survival probability in early and middle life, followed by rapid decline in survival rate in later life. They are typical for species that produce few offspring but care for them well, e.g., humans. In Type III curves, greatest mortality is experienced in early life with relatively low rates of death for those surviving. This type of curve is characteristic of species that produce large number of offspring. Oysters, produce vast number of offspring, only a few of which, live to reproduce. However, once they become established and grow into reproductive

individuals, their mortality rate is extremely low, so they have type III survivorship curve. Type II curves are an intermediate between types I and III, where roughly constant mortality rate and survival probability is experienced regardless of age, e.g., Hydra. 3. (i) The given processes P and Q belong to ELISA (enzyme-linked immunosorbent assay). P is indirect ELISA and Q corresponds to sandwich ELISA. The assay makes use of an enzyme that reacts with a colourless substrate to produce a coloured product. The enzyme is covalently linked to a specific antigen/antibody that recognizes a target antibody/antigen. If the target is present, the antibody-enzyme complex will bind to it and, on addition of the substrate, the enzyme will catalyse the reaction, generating the coloured product. Thus, the presence of the coloured product indicates the presence of the antigen or antibody. (ii) The indirect ELISA is used to detect the presence of antibody and is the basis of the test for HIV infection. The HIV test detects the presence of antibodies that recognize viral core protein antigens. Viral core proteins are adsorbed to the bottom of a well. Antibodies from the person being tested are then added to the coated well. Only a person infected with HIV will have antibodies that bind to the antigen. Finally, enzymelinked antibodies to human antibodies are allowed to react in the well and unbound antibodies are removed by washing. Substrate is then applied. An enzyme reaction yielding a coloured product suggests that the enzyme-linked antibodies were bound to human antibodies, which in turn implies that the patient has antibodies to the viral antigen. Moreover, this assay is quantitative i.e., the rate of the colour-formation reaction is proportional to the amount of antibody originally present. The sandwich ELISA is used to detect antigen rather than antibody. Antibody to a particular antigen is first adsorbed to the bottom of a well. Next, solution containing the antigen (such as blood or urine, in medical diagnostic tests) is added to the well and binds to the antibody. Finally, a second, different antibody to the antigen is added. In this case, the rate of colour formation is directly proportional to the amount of antigen present. Consequently, it permits MT BIOLOGY

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the measurement of small quantities of antigen. 4. (i) Heat loss or gain is directly related to exposed surface area. Shorter extremities are an adaptation to reduce exposed surface area in order to reduce heat loss in cold regions. Therefore, arctic hare (Lepus arcticus) found in tundra region, has smallest external ear size as small size of the ears helps to reduce heat loss in cold conditions. (ii) Given figure shows the North American hares, in which there is a good correlation between external ear size and average temperature experienced by different species. Here, populations of North American hares from different latitudes follow Allen's rule. Increasing latitude indicates decreasing temperatures. Allen's rule states that "animals of colder areas have shorter extremities (leg tail, ears, feet) as compared to animals of warmer areas". 5.(i) Structure labelled A is b-pleated secondary structure. In b -pleated secondary structure, two or more polypeptide chains get interconnected by hydrogen bonds. A sheet is produced, therefore this structure

is called pleated sheet or b-pleated sheet. Adjacent strands of polypeptides may run in same direction or in opposite directions. The adjacent strands are held together by hydrogen bonds formed between –CO and –NH groups to form sheet. (ii) The differences between enzyme cellulase and its substrate cellulose are: Cellulose

Cellulase

1. Made up of amino Made up of b-glucose. acids. 2. Peptide bond between Glycosidic bonds (b1-4) monomers. between monomers. 3. Ionic bonds, hydro- These are not present. phobic interactions and disulphide bonds are present. 4. No rotation of amino Alternate b -glucose acids. rotated through 180°. 5. Globular in nature.

Fibrous in nature. nn

Foods to help you concentrate Have you recently felt like you're out of focus? Do you find it difficult to concentrate on your work? Experts say that some foods contain essential vitamins and minerals that help you concentrate better. Here's a list...

AVOCADO Avocados are said to be one of the most effective foods when it comes to improving concentration. They contain monounsaturated fats that boost the nerves in your brain, which are responsible for taking in information and registering facts and events.

GREEN VEGETABLES While green leafy veggies are good for overall health, they are exceptionally good for your brain. Spinach, broccoli, iceberg lettuce, kale and cabbage are packed with antioxidants as well as B vitamin folate, 30

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which will improve your ability to focus better.

PEANUTS This humble nut may seem like a not-so-important snack but studies say that peanuts contain fibre and a host of minerals and vitamins that help you think more clearly and focus better.

DRY FRUITS Have a handful of dry fruits daily, and see how much they benefit you. If you don't want to eat them plain, add them to your breakfast cereal. They will increase energy levels and help you pay more attention to daily tasks.

BLUEBERRIES Add blueberries to your snacks, smoothies or breakfast regularly. The fibre content in them will make sure your attention doesn't waver. And since they're low cal, you needn't worry about any weight gain either. Courtesy : The Times of India

UnIT-VII : GEnETICS AnD EVOLUTIOn ChApTEr-5 : prInCIpLES OF InhErITAnCE AnD VArIATIOn Multiple Choice Questions 1. If both parents are carriers for thalassaemia, which is an autosomal recessive disorder, what are the chances of pregnancy resulting in an affected child? (a) 25% (b) 100% (c) No chance (d) 50% 2. A child’s blood group is ‘O’. His parents’ blood groups cannot be (a) A and B (b) A and A (c) AB and O (d) B and O. 3. Down’s syndrome in humans is due to (a) three ‘X’ chromosomes (b) three copies of chromosome 21 (c) monosomy (d) two ‘Y’ chromosomes. 4. If a cross between two individuals produces offsprings with 50% dominant character (A) and 50% recessive character (a), the genotypes of parents are (a) Aa × Aa (b) Aa × aa (c) AA × aa (d) AA × Aa. 5. Which one of the following conditions correctly describes the manner of determining the sex? (a) Homozygous sex chromosomes (ZZ) determine female sex in birds.

(b) XO type of sex chromosomes determine male sex in grasshopper. (c) XO condition in humans as found in Turner’s syndrome, determines female sex. (d) Homozygous sex chromosomes (XX) produce male in Drosophila. 6. In a typical Mendelian dihybrid cross, one parent is homozygous for both dominant traits and another parent is homozygous for both recessive traits. In the F2 generation , both parental combinations and recombinations appear. The phenotypic ratio of parental combinations to recombinations is (a) 10 : 6 (b) 12 : 4 (c) 9 : 7 (d) 15 : 1. 7. The existence of non-beneficial alleles in heterozygous genotype within a population is (a) genetic load (b) genetic drift (c) genetic flow (d) selection. 8. Due to nondisjunction of chromosomes during spermatogenesis, some sperms carry both sex chromosomes (22A + XY) and some sperms do not carry any sex chromosome (22A + O). If these sperms fertilise normal eggs (22A + X), what types of genetic disorders appear among the offsprings? (a) Klinefelter’s syndrome and Turner’s syndrome (b) Down’s syndrome and Klinefelter’s syndrome (c) Down’s syndrome and Turner’s syndrome (d) Down’s syndrome and Cri-du-chat syndrome MT BIOLOGY

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9. Which of the following genotypes of man shows presence of one Barr body? (a) XY (b) XXXY (c) XXY (d) All of these 10. Which of the following is not a X-linked recessive disease? (a) Haemophilia (b) Colour blindness (c) b-thalassaemia (d) Glucose-6-phosphate dehydrogenase deficiency 11. Haemophilia is a genetic disorder, in which (a) blood fails to coagulate after an injury (b) there is delayed coagulation of blood (c) blood clots in blood vessels (d) blood cell count falls. 12. Given diagram shows a pair of homologous chromosomes during meiosis.

A plant of type ‘H’ will produce seeds with the genotype identical to seeds produced by the plants of (a) Type M (b) Type J (c) Type P (d) Type N. 15. A man having the genotype EEFfGgHH can produce P number of genetically different sperms, and a woman of genotype IiLLMmNn can generate Q number of genetically different eggs. Determine the values of P and Q. (a) P = 4, Q = 4 (b) P = 4, Q = 8 (c) P = 8, Q = 4 (d) P = 8, Q = 8

True or False 16. When I and I B are present together, and both of them express their own types of proteins, it is called co-dominance. A

17. Pedigree analysis provides a strong tool, which is utilized to trace the inheritance of a specific person. Maximum crossing over will occur between genes (a) A and a, D and d (b) C and d, c and D (c) B and c, b and C (d) A and d, a and D. 13. Wife is PTC non-taster and husband is PTC taster. Their son is taster but daughters are non-tasters. This is not a sex linked trait. Which pedigree is correct?

19. Female butterflies have one Z and one W chromosome, whereas the males have a pair of Z chromosomes, besides the autosomes. 20. Deletions and insertions of base pairs of DNA cause frame-shift mutations.

(a)

(b)

21. Sickle cell anaemia is caused by substitution of valine (Val) by glutamic acid (Glu) at the sixth position of alpha globin chain of haemoglobin.

(c)

(d)

22. Physical, psychomotor and mental development is retarded in individuals suffering from trisomy of 21st chromosome.

14. The given Punnett’s square represents the pattern of inheritance in a dihybrid cross where yellow (Y) and round (R) seed condition is dominant over white (y) and wrinkled (r) seed condition.

32

18. Point mutations are commonly observed in cancer cells.

YR

Yr

yR

yr

YR

F

J

N

R

Yr

G

K

O

S

yR

H

L

P

T

yr

I

M

Q

U

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23. Heterozygotes always express a phenotype which is a blend of both the homozygous parental phenotypes. 24. Punnett square is a graphical representation to calculate probability of all possible genotypes of offspring. 25. Mendel’s approach of using mathematics to explain biological phenomena was totally new and acceptable to most biologists.

Match The Columns 26. Match Column-I with Column-II. Column-I Column-II A. Male heterogamety (i) 13-Trisomy B. Patau’s syndrome (ii) Triticale C. Allopolyploidy (iii) 18-Trisomy D. Autoallopolyploidy (iv) Drosophila E. Turner’s syndrome (v) Helianthus F. Female heterogamety (vi) Monosomy G. Edward’s syndrome (vii) ZW-ZZ system 27. Match Column-I with Column-II. (There can be more than one match for items in Column I). Column-I Column-II A. Sex linked disorder (i) Sickle cell anaemia B. Translocation (ii) Haemophilia C. Deletion (iii) Turner’s syndrome D. Autosomal disorder (iv) Ancon sheep E. Discontinuous (v) Cystic fibrosis variation F. Continuous variation(vi) Colour blindness G. Heteroploidy (vii) Skin colour (viii) Polydactyly (ix) Klinefelter’s syndrome (x) Length of fingers (xi) Cri-du-chat syndrome (xii) Chronic myeloid leukemia (xiii) Philadelphia chromosome (xiv) Intrachromosomal aberration

passage Based Questions 28.(A) Complete the given passage with appropriate words or phrases. (a) The male gametes, in humans, are of two types. The (22 + X) sperms are called (i) , while (22 + Y) sperms are called (ii) . As Y chromosome determines male sex of the embryo, it is called (iii). (b) The (iv) gene on Y chromosome brings about differentiation of embryonic gonads into testes. In its absence, gonads differentiate into ovaries after (v) week of embryonic development, resulting in female sex, which is therefore called (vi) sex.

(B)

Read the given passage and correct the errors, wherever present. Morgan observed during oogenesis in a few insects, a structure called Y body, which was later named Y chromosome. It is an autosome. Presence of one extra copy of it in a male individual leads to Turner’s syndrome.

Assertion & reason In each of the following questions, a statement of Assertion (A) is given and a corresponding statement of Reason (R) is given just below it. Of the statements, mark the correct answer as : (a) if both A and R are true and R is the correct explanation of A (b) if both A and R are true but R is not the correct explanation of A (c) if A is true but R is false (d) if both A and R are false. 29. Assertion : Environment easily influences the phenotypic expression of polygenic inheritance. Reason : Polygenic inheritance is related to both quantitative and qualitative inheritance. 30. Assertion : In F2 generation of Mendelian crosses, 1/2 the individuals showed genotype of F1 generation. Reason : In such crosses, the F1 individuals incompletely expressed the dominant trait. 31. Assertion : Trihybrid cross involves inheritance of three unit factors controlling three distinct characters. Reason : The phenotypic trihybrid ratio in F2 generation is 27 : 9 : 9 : 9 : 9 : 3 : 3 : 1. 32. Assertion : The basic principles of inheritance, as formulated by Mendel are collectively called Mendel’s principles of inheritance or Mendelism. Reason : Principle of dominance and the principle of segregation are Mendel’s first and second laws, respectively. 33. Assertion : The F2 ratio of a Mendelian dihybrid cross shows two types of recombinants. Reason : Generally, the alleles of the two different characters are free to assort and combine, independent of each other. 34. Assertion : Radiations occur naturally in the form of UV rays, as well as ionic radiations from space. Reason : UV rays produce thiamine dimers, resulting in mutations. MT BIOLOGY

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35. Assertion : Grasshopper is an example of XY type of sex determination, similar to Drosophila. Reason : Males have X and Y chromosomes, while females have a pair of X chromosomes. 36. Assertion : Despite difference in morphology, X and Y chromosomes synapse during zygotene. Reason : The homologous part possesses similar genes, called XY linked genes. 37. Assertion : A male can pass sex linked traits to his grandson through his daughter. Reason : This type of cris-cross inheritance is called dia-andric inheritance. 38. Assertion : Translocation does not add any genes or gene groups. Reason : It results in new linkages and in position effect, resulting in chromosomal aberrations, in some cases.

Figure Based Questions 39. Refer to the given figure showing the pedigree A with shaded symbols. Identify whether the shaded symbol refers to: (i) a dominant or a recessive allele. (ii) an autosomal or a sex linked allele.

ChApTEr-6 : MOLECULAr BASIS OF InhErITAnCE Multiple Choice Questions 1. If the sequence of bases in the coding strand of a double stranded DNA is 5’-GTTCGAGTC-3’, the sequence of bases in its transcript will be (a) 5’-GACUCGAAC-3’ (b) 5’-CAAGCUCAG-3’ (c) 5’-GUUCGAGUC-3’ (d) 5’-CUGAGCUUG-3’. 2. The result of which of the following reaction experiments carried out by Avery et al on Streptococcus pneumoniae has proved conclusively that DNA is the genetic material? (a) Live ‘R’ strain + DNA from ‘S’ strain + RNAase (b) Live ‘R’ strain + DNA from ‘S’ strain + DNAase (c) Live ‘R’ strain + Denatured DNA of ‘S’ strain + Protease (d) Heat killed ‘R’ strain + DNA from ‘S’ strain + DNAase 3. Which of the following events would occur in ‘lac operon’ of E. coli when the growth medium has high concentration of lactose? (a) The repressor protein attaches to the promoter sequence and derepresses the operator. (b) The structural genes fail to produce polycistronic mRNA. (c) The inducer molecule binds to repressor protein and RNA polymerase binds to promoter sequence. (d) The repressor protein binds to RNA polymerase and prevents translation. 4. The figure gives an important concept in the genetic implication of DNA.

40. Observe the pedigree A again, and give its genotype in pedigree B.

Fill the blanks A, B and C. (a) A-Maurice Wilkins, B-Transcription, C-Translation (b) A-James Watson, B-Replication, C-Extension (c) A-Erwin Chargaff, B-Translation, C-Replication (d) A-Francis Crick, B-Translation, C-Transcription 5. Select the correct statement regarding protein synthesis. (a) When the small subunit of the ribosome encounters an mRNA the process of translation begins. (b) Peptidase catalyses the formation of peptide bond.

34

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(c) UTRs are present between the start codon and stop codon. (d) At the end of translation, the release factor binds to the initiation codon. 6. Read the following statements and choose the correct option. A. Nitrogenous base is linked to the pentose sugar through a N-glycosidic linkage. B. Phosphate group is linked to 5’-OH of a nucleoside through phosphoester linkage. C. Two nucleosides are linked through 3’ – 5’ N-glycosidic linkage. D. Negatively charged DNA is wrapped around positively charged histone octamer to form nucleosome. E. The chromatin that is more densely packed and stains dark is called euchromatin. (a) A, B and C alone are wrong. (b) D alone is wrong. (c) C and E alone are wrong. (d) A alone is wrong. 7. Which one of the following is correct? (a) Introns are present in mRNA and exons are present in tRNA. (b) Codons are present in mRNA and anticodons in tRNA. (c) Every intron is a set of three terminator codons. (d) Exons are present in eukaryotes while introns are present in prokaryotes. 8. Match the codons with their respective amino acids and choose the correct answer. Codons Amino acids A. UUU 1. Serine B. GGG 2. Methionine C. UCU 3. Phenylalanine D. CCC 4. Glycine E. AUG 5. Proline (a) A – 3, B – 4, C – 1, D – 5, E – 2 (b) A – 3, B – 1, C – 4, D – 5, E – 2 (c) A – 3, B – 4, C – 5, D – 1, E – 2 (d) A – 2, B – 4, C – 1, D – 5, E – 3 9. A mixture containing DNA fragments, A, B, C and D, with molecular weights of A + B = C, A > B and D > C, was subjected to agarose gel electrophoresis. The positions of these fragments farthest from cathode and nearest to anode sides of the gel would be (a) B, A, C, D (b) A, B, C, D (c) C, B, A, D (d) B, A, D, C.

10. Which DNA molecule among the following will melt at lowest temperature ? (a) 5′ - A- A-T- G-C-T- G-C-3′ 3′ - T -T- A-C-G-A-C-G-5′ (b) 5′- A-A-T-A-A-A-G-C-T-3′ 3′- T- T- A-T- T- T- C-G-A-5′ (c) 5′-G-C-A-T- A-G-C-T-3′ 3′-C-G-T- A-T-C-G-A-5′ (d) 5′- A-T- G-C- T- G-A-T-3′ 3′- T- A-C-G-A- C- T- A-5′ 11. Given below are the steps of protein synthesis. Arrange them in correct sequence and select the correct option. (i) Codon-anticodon reaction between mRNA and aminoacyl tRNA complex. (ii) Attachment of mRNA and smaller sub-unit of ribosome. (iii) Charging or aminoacylation of tRNA. (iv) Attachment of larger sub-unit of ribosome to the mRNA-tRNAMet complex. (v) Linking of adjacent amino acids. (vi) Formation of polypeptide chain (a) (ii) → (i) → (iii) → (v) → (iv) → (vi) (b) (v) → (ii) → (i) → (iii) → (iv) → (vi) (c) (iii) → (ii) → (iv) → (i) → (v) → (vi) (d) (iii) → (ii) → (i) → (iv) → (v) → (vi) 12. Arrange the various steps of DNA fingerprinting technique in the correct order. (i) Separation of DNA fragments by electrophoresis (ii) Digestion of DNA by restriction endonucleases (iii) Hybridisation using labelled VNTR probe (iv) Isolation of DNA (v) Detection of hybridised DNA fragments by auto radiography (vi) Transferring the separated DNA fragments to nitrocellulose membrane (a) (iv) → (ii) → (i) → (vi) → (iii) → (v) (b) (iv) → (i) → (ii) → (iii) → (vi) → (v) (c) (ii) → (i) → (iv) → (vi) → (iii) → (v) (d) (iii) → (v) → (iv) → (ii) → (i) → (vi) 13. If Meselson and Stahl’s experiment is continued for four generations in bacteria, the ratio of 15N/15N: 15N/14N: 14N/14N containing DNA in the fourth generation would be (a) 1 : 1 : 0 (b) 1 : 4 : 0 (c) 0 : 1 : 3 (d) 0 : 1 : 7. MT BIOLOGY

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14. DNA is a polymer of nucleotides which are linked to each other by 3’-5’phosphodiester bond. To prevent polymerisation of nucleotides, which of the following modifications would you choose? (a) Replace purine with pyrimidines (b) Remove/Replace 3’-OH group in deoxyribose (c) Remove/Replace 2’-OH group with some other group in deoxyribose (d) Both (b) and (c) 15. The differences between mRNA and tRNA are that (i) mRNA has more elaborated 3-dimensional structure due to extensive base pairing. (ii) tRNA has more elaborated 3-dimensional structure due to extensive base pairing. (iii) tRNA is usually smaller than mRNA. (iv) mRNA bears anticodon but tRNA has codons. (a) (i) and (ii) (b) (ii) and (iii) (c) (i), (ii) and (iii) (d) (i), (ii), (iii) and (iv)

True or False 16. The string-on-beads structure in chromatin is packaged to form chromatin fibres, that are further condensed at metaphase stage to form chromatids. 17. The negatively charged DNA is wrapped around the positively charged histone octamer to form a nucleotid. 18. The DNA dependent DNA polymerase catalyses polymerisation only in one direction, that is 5′ → 3′. 19. The structural gene in a transcription unit is monocistronic in prokaryotes and polycistronic in eukaryotes. 20. The coding strand has 5′ → 3′ polarity and does not code for anything. 21. Many times in the eukaryotic cell, the translation can begin much before the mRNA is fully transcribed. 22. The chemical method developed by Har Gobind Khorana was instrumental in synthesising RNA molecules with defined combinations of bases. 23. Untranslated regions or UTRs are present at 5′ end (before start codon) only. 24. Chromosome 1 has the maximum number of genes, 2968, while chromosome Y has the least number of genes, about 231. 36

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25. Polymorphism arises due to mutations and is the basis of genetic mapping of human genome and DNA fingerprinting.

Match The Columns 26. Match Column-I with Column-II. Column-I Column-II A. Z gene (i) Clover leaf B. tRNA (ii) 5′ → 3′ C. Promoter (iii) hnRNA D. Leading strand (iv) Chaperones E. Lagging strand (v) 3′ → 5′ F. Primary transcript (vi) TATA box G. Helix formation (vii) b-gal 27. Match Column-I with Column-II. (There can be more than one match for items in Column-I). Column-I Column-II A. Initiation codon (i) AUG B. Phenylalanine (ii) UAA C. Template strand (iii) UUU D. Termination codon (iv) Minus strand E. Nontemplate strand (v) Plus strand F. Arginine (vi) GUG G. Trp operon (vii) UGA (viii) AGG (ix) Antisense strand (x) CGU (xi) UUC (xii) Sense strand (xiii) Corepressor (xiv) Negative control

passage Based Questions 28.(A) Complete the given passage with appropriate words or phrases. Transcription requires (i) polymerase and (ii) factors in eukaryotes. There are at least three RNA polymerases in eukaryotes. (iii) is for transcribing tRNA, 5SRNA and some snRNAs; (iv) is for transcribing mRNA and snRNAs while (v) is for rRNA, except 5S rRNA. (B) Read the passage and correct the errors, wherever present. Additional nucleotides are added to ends of RNAs for specific functions — CCC segment in tRNA, cap nucleotides at 3′ end of mRNA and poly-G segments at 5′ end of mRNA. The cap is formed by modification of ATP into 6-methyl adenosine or 6mA.

Assertion & reason In each of the following questions, a statement of Assertion (A) is given and a corresponding statement of Reason (R) is given just below it. Of the statements, mark the correct answer as : (a) if both A and R are true and R is the correct explanation of A (b) if both A and R are true but R is not the correct explanation of A (c) if A is true but R is false (d) if both A and R are false. 29. Assertion : DNA replication in prokaryotes is semiconservative, semicontinuous and bidirectional. Reason : DNA replication in eukaryotes is semiconservative, bidirectional and continuous.

Reason : Nine tenth of genes are identical to that of the mouse. 37. Assertion : VNTRs are ‘variable number tandem repeats which are inherited by an individual from his parents. Reason : They are also called ‘microsatellites’ and are used as genetic markers in personal identity test. 38. Assertion : A peptide bond (–CO–NH–) is established between the amino group (–NH2) of amino acid at P site, and carboxyl group (–COOH) of amino acid at A-site. Reason : Peptide bond formation during translation is catalysed by peptidyl transferase, a protein enzyme.

Figure Based Questions 39. Refer the given figure and identify the labelled parts P, Q, R, S and T.

30. Assertion: The primary transcript of mRNA is called heterogeneous nuclear RNA or hnRNA. Reason : hnRNA is often larger than the functional RNA. 31. Assertion : Codon is found in both DNA and mRNA. Reason : Codon determines the position of a gene in the DNA. 32. Assertion : Both RNA and DNA are feulgen negative. Reason : Base pairing through hydrogen bonds occurs only in DNA and never in RNA. 33. Assertion : A lot of energy is consumed during protein synthesis. Reason : For every single amino acid incorporated in the peptide chain, 2 ATP and 1GTP molecules are used. 34. Assertion : Out of the two strands of DNA, only one strand is effective in producing mRNA in a given cistron. Reason : It is called antisense strand. 35. Assertion : Aporepressor is a proteinaceous substance

synthesised by regulator gene. Reason : By itself, the aporepressor is unable to block the working of operator gene. 36. Assertion : Human gene count is about the same, as that of the mouse.

40. Fill in the blanks using the letters assigned to figure in previous question. (a) _____ has 7 bases out of which 3 bases form anticodon or nodoc, for attaching to codon of mRNA. (b) _____ lies at 3’ end, opposite the anticodon. (c) _____ has 7 bases and forms site for attaching to ribosome. (d) _____ is the largest loop and is binding site for aminoacyl synthetase enzyme. (e) _____ is not present in all tRNAs, and its exact role is not known. MT BIOLOGY

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ChApTEr-7 : EVOLUTIOn Multiple Choice Questions 1. In a population of 1000 individuals, 360 belong to genotype AA, 480 to Aa and the remaining 160 to aa. Based on this data, the frequency of allele A in the population is (a) 0.4 (b) 0.5 (c) 0.6 (d) 0.7. 2. Choose the wrong statement. (a) Louis Pasteur demonstrated that life comes only from pre-existing life. (b) S.L. Miller observed that electric discharge in a flask containing CH4, H2, NH3 and water vapour at 800° C formed amino acids. (c) Flippers of penguins and dolphins are examples of homology. (d) Analogous structures are the result of convergent evolution. 3. According to Darwin, organic evolution is mainly due to (a) competition within closely related species (b) reduced feeding efficiency in one species due to the presence of interfering species (c) intraspecific competition (d) interspecific competition. 4. Random unidirectional change in allele frequencies that occurs by chance in all populations and especially in small populations is known was (a) migration (b) natural selection (c) genetic drift (d) mutation. 5. The extinct human ancestor, whose fossil was discovered by Edward Lewis from Pliocene rocks of Shivalik Hills of India was (a) Ramapithecus (b) Australopithecus (c) Dryopithecus (d) Homo erectus.

7. Populations are said to be allopatric when (a) they are physically isolated by natural barriers (b) they are sharing the same area but cannot interbreed (c) they live together and breed freely to produce viable offspring (d) they are isolated but often come together for breeding. 8. The first life on earth consisted of (a) provirus (b) protovirus (c) virus (d) bacteria. 9. Adaptive radiation refers to (a) evolution of different species from a common ancestor (b) migration of members of a species to different geographical areas (c) power of adaptation in an individual to a variety of environments (d) adaptations due to geographical isolation. 10. An important evidence in favour of organic evolution is the occurrence of (a) homologous and analogous organs (b) homologous and vestigial organs (c) analogous and vestigial organs (d) homologous organs only. 11. Following table shows data on amino acid substitution in the a chain of haemoglobin in four different mammalian species A, B, C and D. On the basis of the data shown in the table, choose the most appropriate evolutionary tree from those given below. Comparison of Species

Number of Amino Acid Substitution

A and B

19

B and C

26

A and C

27

D and C

27

A and D

20

D and B

1

6. Which one of the following options gives one correct example each of convergent evolution and divergent evolution? Convergent Divergent evolution evolution (a) Eyes of octopus and mammals (b) Thorns of Bougainvillea and tendrils of Cucurbita (c) Bones of forelimbs of vertebrates (d) Thorns of Bougainvillea and tendrils of Cucurbita 38

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Bones of forelimbs of vertebrates Wings of butterflies and birds Wings of butterfly and birds Eyes of octopus and mammals.

(a)

(c)

(b)

(d)

12. The prebiotic atmosphere of the earth was of a reducing nature. It was transformed into an oxidising atmosphere of present day due to the emergence of (a) cyanobacteria (b) angiosperms (c) photosynthetic bacteria (d) eukaryotic algae.

15. Following is the diagrammatic representation of the operation of natural selection on different traits. Which of the following options correctly identifies all the three graphs A, B and C?

13. In a long term experiment on a population of Drosophila melanogaster, the frequency of two alleles ‘a’ and ‘b’ of a multi-allelic locus X over time has been shown in the following graph.

6 students were asked to evaluate the observed pattern and their inferences are given below. Student 1 : Environment is not uniformly selective. Student 2 : Population may be under artificial selection. Student 3 : Genetic variability is progressively reduced. Student 4 : Genetic variability is progressively increased. Student 5 : Mechanism such as genetic drift is operating from time to time. Student 6 : Selection is favouring a particular genotype through directional selection. The appropriate conclusions were drawn by (a) Students 2, 5 and 6 (b) Students 1, 3 and 5 (c) Students, 2, 3 and 6 (d) Students 1, 3 and 6. 14. An isolated population of humans with approximately equal numbers of blue-eyed and brown-eyed individuals was decimated by an earthquake. Only a few brown-eyed people remained to form the next generation. This kind of change in the gene pool is called (a) Hardy-Weinberg equilibrium (b) blocked gene flow (c) bottle-neck effect (d) gene migration.

(a) (b) (c) (d)

A Directional Stabilising Disruptive Directional

B Stabilising Directional Stabilising Disruptive

C Disruptive Disruptive Directional Stabilising

True or False 16. Life appeared almost 4 billion years ago, about 500 milion years after formation of earth. 17. Eyes of octopus and mammals represent homology and convergent evolution. 18. Homologous organs show adaptive radiation. 19. When convergent evolution is found in distantly related species, it is called parallel evolution. 20. Lamarckism and Darwinism, both support that acquired characters, useful to the possessor can be inherited. 21. The same type of mutations can appear in a number of individuals of a species. 22. Both birds and mammals are believed to have evolved from a reptilian ancestor. 23. Modern apes are believed to be the direct ancestors of man. 24. The probability that an allele p with a frequency of A appears on both chromosomes of a diploid individual is simply the product of probabilities, i.e, p2. 25. The blood groups A, B, AB and O are found in both humans and apes, and not in monkeys. MT BIOLOGY

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Match The Columns 26. Match Column-I with Column-II. Column-I Column-II A. Discontinuous (i) Darwin’s finches distribution B. Raymond Dart (ii) Peppered moth C. Hugo de Vries (iii) Charles Darwin D. Adaptive radiation (iv) Tuang baby E. Donald Johnson (v) Saltation F. Industrial melanism (vi) Lung fishes G. Gemmule (vii) Lucy 27. Match Column-I with Column-II. (There can be more than one match for items in Column-I). Column-I Column-II A. Premating isolation (i) Temporal isolation B. Cretaceous (ii) First modern birds C. Microfossils (iii) Origin of reptiles D. Co-evolution (iv) Hybrid inferiority E. Postmating isolation (v) Palynofossils F. Carboniferous (vi) Prey-predator G. Ordovician (vii) Host-pathogen (viii) Fossil fuels (ix) First land plants (x) First monocots (xi) Ethological isolation (xii) Hybrid breakdown (xiii) Rise of gymnosperms (xiv) First jawless fishes

passage Based Questions 28.(A) Complete the given passage with appropriate words or phrases.

(B)

40

When two species are morphologically almost identical, but do not normally interbreed, such species are called (i) , for instance (ii) and (iii) are two species of fruit fly, which do not cross fertilise. Species restricted to a specific area are called (iv) while (v) refers to species in adjacent geographical areas, meeting in narrow regions of overlap. Read the passage and correct the errors, wherever present. Nearctic realm covers Central and South America while Neotropical realm covers Canada and United States MT BIOLOGY

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of America. Ethiopian realm includes Sahara Desert of Africa and Siberia while Palaearctic realm includes Arabia and Madagascar. Palaearctic and Nearctic realm together form Wallace’s line while Holoarctic region lies between Oriental and Australian realm.

Assertion & reason In each of the following questions, a statement of Assertion (A) is given and a corresponding statement of Reason (R) is given just below it. Of the statements, mark the correct answer as : (a) if both A and R are true and R is the correct explanation of A (b) if both A and R are true but R is not the correct explanation of A (c) if A is true but R is false (d) if both A and R are false. 29. Assertion : As oxygen accumulated in the atmosphere, the ultraviolet light changed some of the oxygen into ozone. Reason : The ozone formed a layer in the atmosphere, blocking the UV rays. 30. Assertion: Due to rapid increase in the number of heterotrophs, the nutrients from sea water began to disappear and gradually got exhausted. Reason : This led to the evolution of first anaerobic photoautotrophs and later aerobic photoautotrophs. 31. Assertion : Genetic drift is an evolutionary force. Reason : It occurs only in small isolated populations. 32. Assertion : Disruptive evolution favours both smallsized and large-sized individuals, and eliminates those with mean expression. Reason : It is very common in nature and is also the most important for bringing evolutionary change. 33. Assertion : Balanced polymorphism occurs when different forms coexist in the same population in a stable environment. Reason : In humans, the existence of A, B, AB and O blood groups represent balanced polymorphism. 34. Assertion : Fossil ostracoderms probably evolved from unarmoured ancestors, such as Jamoytius. Reason : Before extinction, ostracoderms gave rise

to the first bony fishes, placoderms and cartilaginous fishes. 35. Assertion : Cro-Magnon is believed to be more

intelligent and cultured than the man of today. Reason : It had a cranial capacity, somewhat more than ours, being about 1650cc. 36. Assertion : Random changes in the allele frequencies of a population, occurring only by chance, constitute genetic drift. Reason : Two important examples of genetic drift are founder effect and gene flow. 37. Assertion : In case of vegetatively propagating plants, somatic cells produce germ cells. Reason : This is against Weismann’s theory of continuity of germplasm. 38. Assertion : Finches of Galapagos Islands influenced Darwin to think about evolutionary change. Reason : He called these birds as Darwin’s finches.

Figure Based Questions 39. Refer to the diagrammatic representation of evolutionary history of vertebrates and identify the labelled parts - A, B, C, D, E, F and G.

40. Fill in the blanks using the letters assigned above. (a) Periods (i) and (ii) belong to coenozoic era, while (iii) , (iv) , (v) periods belong to mesozoic era and (vi) , (vii) periods belong to palaeozoic era. (b) (i) period is called age of man and herbs, while (ii) period is called age of dinosaurs and cycads. (iii) period is called age of mammals, birds, angiosperms, while (iv) is called age of amphibians, ferns and coal forests.

SOLUTIOnS CHAPTER-5 : PRINCIPlES of INHERITANCE AND vARIATIoN

1. (a) 6. (a) 11. (a)

2. (c) 7. (a) 12. (d)

3. (b) 4. (b) 8. (a) 9. (c) 13. (a) 14. (d)

5. (b) 10. (c) 15. (b)

16. false. I A and I B express their own types of sugars not proteins. 17. false. Pedigree analysis is utilised to trace the inheritance of a specific trait, abnormality or disease in multiple generations. 18. false. Chromosomal aberrations are commonly

observed in cancer cells. 19. True

20. True

21. false. It is caused by substitution of glutamic acid (Glu) by valine (Val) at the sixth position of beta globin chain of haemoglobin. 22. True

false. Alleles do not blend although in a few cases of incomplete dominance an intermediate phenotype appears. Heterozygotes express dominant phenotype when there is a complete dominant recessive allele interaction or both the phenotypes when alleles are codominant. 24. True 25. false. It was unacceptable to most biologists. 26. A-(iv), B-(i), C-(ii), D-(v), E-(vi), F-(vii), G-(iii) 27. A-(ii, vi), B-(xii, xiii), C-(xi, xiv), D-(i, v), E-(iv, viii), F-(vii, x), G-(iii, ix) 28.(A) (a) (i) androsperms, (ii) gynosperms, (iii) androsome (b) (iv) TDF, (v) sixth, (vi) default (B) Morgan Henking observed during oogenesis spermatogenesis in a few insects, a structure called Y X body, which was later named Y X chromosome. It is an autosome allosome. Presence of one extra copy of it in a male individual leads to Turner’s Klinefelter’s syndrome. 23.

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29. (c) 30. (c) 31. (d) 32. (c) 33. (a) 34. (c) 35. (d) 36. (a) 37. (c) 38. (b) 39. (i) It is a dominant allele. Also, one parent (father) must be homozygous dominant, while the other parent (mother) must be homozygous recessive. (ii) Since the shaded symbol appears in all offspring irrespective of sex, it refers to an autosomal allele, and not sex linked.

29. 34. 39.

40.

40.

segments at 5′ 3′ end of mRNA. The cap is formed by modification of ATP GTP into 6-methyl adenosine 7-methyl guanosine or 6mA 7mG. (d) 30. (b) 31. (c) 32. (d) 33. (c) (b) 35. (b) 36. (b) 37. (c) 38. (d) P-Anticodon loop Q-T y C loop R-DHU loop S-Variable arm T-Amino acid attaching site (a) P; (b) T; (c) Q; (d) R; (e) S

CHAPTER-7 : EvoluTIoN

1. 6. 11. 16.

CHAPTER-6 : MolECulAR BASIS of INHERITANCE

1. (c) 2. (b) 3. (c) 4. (d) 5. (a) 6. (c) 7. (b) 8. (a) 9. (a) 10. (b) 11. (d) 12. (a) 13. (d) 14. (b) 15. (b) 16. false. It is the beads-on-string structure in chromatin, packaged to form chromatin fibres, that are further coiled and condensed at metaphase stage, to form chromosomes.

false. The negatively charged DNA is wrapped around the positively charged histone octamer to form a nucleosome. 18. True 19. false. It is monocistronic in eukaryotes and polycistronic in prokaryotes. 20. True 21. false. The transcription and translation can be coupled in prokaryotes not eukaryotes as the required machinery for transcription is in nucleus while that for translation is in cytoplasm. 22. True 23. false. UTRs are present at both 5’ end and 3’ end, and are required for efficient translation process. 24. True 25. True 26. A-(vii), B-(i), C-(vi), D-(ii), E-(v), F-(iii), G-(iv) 27. A-(i, vi), B-(iii, xi), C-(iv, ix), D-(ii, vii), E-(v, xii), F-(viii, x), G-(xiii, xiv) 28.(A) (i) DNA dependent RNA (ii) Transcription (iii) Pol III (iv) Pol II (v) Pol I or Pol A (B) Additional nucleotides are added to ends of RNAs for specific functions - CCC CCA segment in tRNA, cap nucleotides at 3′ 5′ end of mRNA or poly-G poly-A 17.

42

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(c) 2. (c) 3. (c) 4. (c) 5. (a) (a) 7. (a) 8. (b) 9. (a) 10. (b) (d) 12. (a) 13. (c) 14. (c) 15. (b) false. Life appeared 500 milion years after the formation of earth, 4 billion years back. 17. false. They represent analogy. 18. True 19. false. When convergent evolution is between closely related species, it is parallel evolution. 20. True 21. True 22. True 23. false. No modern ape is director ancestor, but it is believed that man and modern apes are cousins descended from common ancestors. 24. false. It is the probability of allele A with frequency p. 25. false. Only blood groups A and B are found in apes and not AB and O. 26. A-(vi), B-(iv), C-(v), D-(i), E-(vii), F-(ii), G-(iii) 27. A-(i, xi), B-(ii, x), C-(v, viii), D-(vi, vii), E-(iv, xii), F-(iii, xiii), G-(ix, xiv) 28.(A) (i) Sibling species, (ii) Drosophila pseudoobscura, (iii) Drosophila persimilis, (iv) endemic species, (v) parapatric species. (B) Nearctic Neotropical realm covers Central and South America while Neotropical Nearctic realm covers Canada and United States of America. Ethiopian Palaearctic realm includes Sahara Desert of Africa and Siberia while Palaearctic Ethiopian realm, includes Arabia and Madagascar. Palaearctic and Nearctic realm together form Wallace’s line Holoarctic region while Holoarctic region Wallace’s line lies between Oriental and Australian realm. 29. (b) 30. (c) 31. (c) 32. (c) 33. (b) 34. (b) 35. (a) 36. (c) 37. (b) 38. (c) 39. A-Carboniferous; B-Permian; C-Triassic D-Cretaceous; E-Jurassic; F- Quaternary; G-Tertiary 40. (a) (i) F; (ii) G; (iii) D; (iv) E; (v) C; (vi) B; (vii) A (b) (i) F; (ii) E; (iii) G; (iv) A nn

Class XI

Sensory Receptors – – – – – J&K

–

1

– –

– –

– –

1

– K.CET

–

– Kerala

2

2

– 2

1 1

1 –

–

2

4

–

AMU

–

1

AIIMS

2013

1 1

2012 2011

– –

2010

primitive type of sensory receptor is a single unspecialised sensory or afferent neuron with its receptor endings meant for detecting stimuli. It is called primary sense cell e.g., olfactory cells. It is also called 1st order neuron. • Secondary sense cells or 2nd order neurons are modified epithelial cells that form synaptic connections, via interneurons with the sensory neurons, and transmit impulses to the cNS e.g., mammalian taste buds. • The most complex sensory receptors, are called sensory organs. They consist of numerous sense cells, sensory neurons (1st order, 2nd order, 3rd order neurons and so on) and associated accessory structures, e.g., eye and ear have a level of complexity of sense organs.

–

touch, taste and olfaction. • all sensory receptors are similar in basic structure. The simplest and most

2009

• The commonly recognized sensory systems are those for vision, hearing,

AIPMT/NEET

from the receptors to the brain (i.e., the ascending or sensory tracts in the spinal cord) and parts of the brain, that deal primarily with processing the information (i.e., somatosensory cortex) in the parietal lobe.

Analysis of various PMTs from 2009-2014

sensory information. It enables us to detect changes within our own body and in our environment. • The information about the changes within the body is used to maintain homeostasis. • a sensory system consists of sensory receptors that receive stimuli from external or internal environment, neural pathway that conducts information

2014

• a sensory system is a part of nervous system responsible for processing

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Working of SenSory receptorS • a stimulus is some form of energy : light, sound,

pressure, heat, osmotic potential, electric current, and chemical changes. • Each type of receptor is sensitive to a specific stimulus and almost non-responsive to other stimuli. Over stimulation of any receptor is painful and virtually all receptor types function as nocireceptors at one time or another e.g., searing heat, extreme cold, excessive pressure and chemicals released at sites of inflammation are all interpreted as painful stimuli. • an animal responds to a stimulus in a four-step process: (i) Sensory transduction : Sensory receptors transduce (transform) the energy of a stimulus into a localized nonpropagated electrical response or generator potential which initiates nerve impulses in the neuron leaving the receptor. (ii) Transmission : The sensory neuron relays the nerve impulse to the brain directly or through the spinal cord. (iii) Integration : Nerve impulses (action potentials, ften called receptor or generator potentials) that reach the brain via sensory neurons are termed sensations. In the brain, the sensations are analysed and interpreted as perceptions. Thalamus is the main, and cerebral cortex is the subsidiary centre of this analysis. The brain transmits motor impulses to appropriate effectorsmuscles or glands. (iv) Response : Effectors produce suitable responses. muscles contract, or glands secrete chemicals, in response to the information sent to the brain by the receptors.

• receptors for touch, heat, cold, light and sound are

examples of exteroceptors. • Bare nerve endings in the skin that detect pain are also exteroceptors.

interoceptors • They are sensory receptors which receive stimuli from

the inner parts, tissue or cells. • They are of two subtypes:

(i) proprioceptors • The proprioceptors are located in the skeletal muscles, joints, tendons, etc. • They given information about movement and position of body. Two types of muscle receptors are muscle spindle and Golgi tendon organ. (ii) Visceroceptors • Visceroceptors are located in the viscera. • They are affected by stimuli originating within the body itself and cause sensations, such as pain, hunger, thirst, fatigue, nausea, sex, etc. • They also monitor blood pressure, carbon dioxide level, body temperature, osmotic relationships, ph, etc. • The visceroceptors are simple and mostly consist of free nerve endings.

claSSification • receptors are classified under two major divisions of

sensory system. These are : (i) General sensory system : comprises of receptors in the skin, joints, skeletal muscles and internal organs. (ii) Special sensory system : comprises of receptors present in highly complex organs like eye, ear, nose and tongue.

classification of receptors according to their location exteroceptors • Exteroceptors are usually located near the surface

of the body, and they detect changes in the external environment. 44

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classification of receptors according to their function • Teloceptors : They are sensitive to changes occurring at a distance, e.g., ear, eye. • Nociceptors : They are receptors sensitive to deep

pain and damage to tissue. • Proprioceptors : Sensitive to changes in tendons, muscles and joint movements. • Labyrinthine receptors : Sensitive to the different positions of body in space.

classification of receptors according to the type of stimuli they receive

These receptors are of five main types: mechanoreceptors, photoreceptors, chemoreceptors, electroreceptors and thermoreceptors, as shown in the table below: Table: Types of receptors Name of receptor

Types and Stimulated by

1. Thermoreceptors respond to alteration in temperature 2. Mechanoreceptors (for mechanical stimuli) stimulated by mechanical deformation like touch, pressure

Examples End bulb of Krause in skin (also called frigidoreceptors) ruffini’s organs in skin (also called caloreceptor)

(a) Tangoreceptors (located in the skin) Touch and pressure (iv) Pacinian corpuscles—for pressure (b) Phonoreceptors Sound waves (c) Statoreceptors acceleration and gravity (d) Algesireceptors (Pain) (e) Proprioreceptors Position of parts of body (f) Rheoreceptors Pressure waves and water currents (g) Baroreceptors Blood pressure

Organ of corti in internal ear. hair cells in cristae and maculae in internal ear Free nerve endings Free nerve endings, neuromuscular and neurotendinous spindles. Lateral line sense organs in fish. Nerve endings in walls of atria, vena cava, carotid sinuses, aortic arch.

3. Photoreceptors (for visual Light wavelengths (electromagnetic) stimuli) stimulated by light

retina in vertebrate eye, Ommatidia in compound eyes of arthropods

4. Chemoreceptors (a) Gustato-receptors stimulated by chemical Taste due to chemicals in solution influences (b) Olfactory receptors (Olfactoreceptors) Smell due to volatile chemicals (c) Humidoreceptors humidity

Taste buds of tongue

5. Electroreceptors

Skin of some fishes.

Effective currents in surrounding water.

taSte receptorS (guStatoreceptorS) • The receptors for taste are found in about 10,000 taste

buds, mostly located on the tongue but also found on the palate, pharynx and epiglottis, and even in the proximal part of oesophagus. • The number of taste buds declines with 45 years of age. • atleast 13 possible chemical receptors are found in

Olfactory epithelium, ampulla of Lorenzini (Scoliodon) Skin

taste cells such as 2 sodium receptors, 2 potassium receptors, 1 chloride receptor, 1 adenosine receptor, 1 inosine receptor, 2 sweet receptors, 2 bitter receptor, 1 glutamate receptor and 1 hydrogen ion receptor. • a person can perceive hundreds of different tastes. They are all supposed to be combination of the elementary taste sensations - sour, salty, sweet, bitter and ‘umami’ or delicious. • On the tongue, taste buds are found in 3 types of papillae - circumvallate, fungiform and foliate papillae. MT BIOLOGY

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Structure of taste bud • Each taste bud is an oval body consisting of three kinds

of cells: (i) Gustatory receptors cells : They bear at the free end microvilli projecting into the taste pore. The microvilli have special protein receptor sites for taste-producing molecules and come in contact with the food being eaten. Nerve fibres of the cranial nerves VII (Facial), IX (Glossopharyngeal) or X (Vagus) end around the gustatory receptor cells, forming synapses with them. The gustatory receptor cells (taste cells) survive only about 10 days and are then replaced by new cells. (ii) Supporting cells : These cells lie between the gustatory receptor cells in the taste bud. They bear microvilli but lack nerve endings. (iii) Basal cells : These cells are found at the periphery of the taste bud. They produce supporting cells, which then develop into gustatory receptor cells.

Working • Specific chemicals in solution, pass into the taste bud

through the taste pore, to come in contact with the protein receptor sites on the microvilli of the gustatory receptor cells. The latter set up nerve impulses in the sensory nerve fibres.

• mammals can locate upto 4000 different odours. • The receptors for smell occur in a small (about 5 cm2)

patch of olfactory neuroepithelium (pseudostratified epithelium) located in the roof of the nasal cavity, with nearly 20 million olfactory receptors. • The free nerve endings of cranial nerve I are located diffusely throughout the nasal respiratory epithelium, including regions of the olfactory neuroepithelium.

Structure of olfactoreceptor • Olfactory

epithelium (also called Schneiderian membrane) is a modified pseudostratified epithelium. It is yellowish in colour and has three types of cells : receptor cells, supporting cells and basal cells, resting on a thick lamina propria. (i) Receptor cells – These are also called olfactory cells, or olfactoreceptors. They act as sensory receptors as well as conducting neurons. They are spindle-shaped bipolar neurons with rounded nuclei in the middle region. Olfactory receptor cells are unique in that they are the only neurons that undergo turnover throughout adult life. The olfactory cells survive only for about 2 months. (ii) Supporting cells – These are columnar cells with large oval nuclei. They lie between the olfactory cells to support them. (iii) Basal cells – These are small cells that do not reach the surface. They give rise to new olfactory cells to replace the worn out ones. • Olfactory glands (Bowman’s glands) – many olfactory glands occur below the olfactory epithelium, that secrete mucus to spread over the epithelium, to keep it moist. The mucus also protects the cells from dust and bacteria.

• The facial nerve (VII) serves the anterior two-thirds

of the tongue, the glossopharyngeal nerve (IX) serves the posterior one-third of the tongue and the vagus nerve (X) serves the pharynx and epiglottis but not the tongue.

Smell receptorS (olfactoreceptorS) • Olfactory sensation is the most primitive of all special

senses and is much more acute than taste, with smell receptor as much as 3,400 times more sensitive than taste receptors. 46

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Working • The dissolved chemicals stimulate the olfactory receptors

by binding to protein receptors in the olfactory hairs

(cilia), membranes and opening specific Na+ and K+ channels. This leads ultimately to an action potential that is conducted to the first relay station in the olfactory bulb. • The fibres of the olfactory nerves synapse with mitral cells (second order neurons) in complex structures called glomeruli (balls of yarn). When the mitral cells are activated, impulses travel from the olfactory bulbs via olfactory tracts to main destinations (e.g., temporal lobe of the cerebrum).

in diameter and consists of tissues present in three concentric layers : – Outermost fibrous layer consists of sclera and cornea. – middle vascular layer (also called uvea) consists of choroid, ciliary body and iris. – Innermost nervous layer consists of retina.

a. fibrous coat

DiSorDerS of Smell anD taSte Anosmia – Loss of the ability to smell. Hyperosmia – Increased sensitivity to odours. Dysosmia – Disagreeable or distorted sense of smell. Hyposmia – Diminished sense of smell. Hypogeusia – Diminished sense of taste. Ageusia – Loss of the sense of taste (Eg. Whales) Dysgeusia – Distorted sense of taste. Sinusitis – an infection of the sinuses (cavities, or airfilled pockets) near the nose. These infections usually occur after a cold or after an allergic inflammation. • Rhinitis – Inflammation of the nasal mucous membrane. • Rhinorrhoea – Nasal flow. • • • • • • • •

Human eye • The organs of sight are a pair of eyes in

human. The eyes are situated in deep protective bony cavities, called the orbits or eye sockets of the skull. The study of structure, function and diseases of the eye is called ophthalmology.

Structure of the eye • human

eyes are spherical structures. Each eye is about 2.5 cm

(i) Sclera • It is an opaque, fibro-elastic collagen capsule that forms the outermost covering. • It is bluish white in appearance, except at the front, where it forms the transparent cornea. • It forms the posterior 5/6th part of the eye ball. • Sclera maintains the shape of the eyeball and protects all the inner layers of the eye. (ii) Cornea • It is a thin transparent, front part of sclera that forms a slight bulge (called conjunctiva) at the front, and covers about 1/6th part of the sclera. • It lacks blood vessels but is rich in nerve endings, and absorbs oxygen from air. • Stimulation of nerve endings reflexly causes blinking and tear flow. • The cornea is kept moist by tears and mucus from conjunctival and lacrimal glands. • as cornea is avascular so its living materials e.g., epithelium, and endothelium get nutrition from aqueous humour, and from super marginal plexus of blood vessels. Therefore, corneal transplant is easiest, with over 90% success rate. MT BIOLOGY

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• cornea allows the light to pass into the eye. Due to its

curvature, it helps in focusing a real inverted image of the object on retina. • The cornea also serves as a filter, screening out some of the most damaging ultraviolet (UV) wavelength in sunlight. • at the junction of sclera and cornea, there is a circular canal of Schlemm. The aqueous humour is drained off into this canal and then into the blood. (iii) conjunctiva • It is a thin, transparent, membranous layer present over the cornea, and is continuous with the corneal epithelium. • It is composed of a stratified epithelium and also lines the eyelids. • The conjunctiva is thin, mildly cornified and richly supplied with free nerve endings. It is nourished by tiny blood vessels that are nearly invisible to the naked eyes. • In the sore or “pink” eyes condition, the conjunctiva gets inflammed, causing conjunctivitis. • Conjunctiva protects the cornea and also secretes oil and mucus that moisten and lubricate the eye.

(iii) iris • It is the anterior part of vascular coat which lies behind the cornea. It is centrally perforated by pupil. • Its pigment gives eye its colour (depending upon the amount of pigment present) like black, blue or green. • In some cases there is no pigment at all, so the eye is light. Albinos lack pigments in the skin, hair and iris. Their pink colour of iris is due to reflection of light from the blood vessels of iris. • The movement of iris controls the size of pupil. The iris contains two sets of smooth muscles : circular muscles or sphincters and radial muscles or dilators, of ectodermal origin. These muscles regulate the amount of light entering the eye ball. The radial muscles contract in dim light, and the circular muscles contract in bright light. • Pupil in woman is larger than that in man.

B. Vascular coat (i) choroid • It is a pigmented layer present beneath the sclera. It is composed of connective tissue and is of dark brown colour. • It is homologous to the pia-arachnoid of the brain. • It contains numerous blood vessels which nourish the retina. The pigmentation prevents reflection within the eye. (ii) ciliary body • The ciliary body extends towards the inside of the eye from the choroid coat. • This part of the eye is less vascular, thick, less pigmented and composed of ciliary muscles and ciliary processes, attached to the lens capsule by suspensory ligaments. • The ciliary processes secrete aqueous humour in the anterior chamber of the eye ball. • Ciliary muscles are a complex set of smooth muscles and are of two types: circular and meridional. These are controlled by autonomic nervous system, and alter the shape of lens. • The contraction of ciliary muscles results in spherical shape of the lens and the relaxation, in the flattened shape (i.e. accommodation). 48

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c. nervous coat : retina • The third and inner coat of the eyeball, the retina

(nervous tunic), lines the posterior three-quarters of the eyeball and is the beginning of the visual pathway. • The retina consists of a pigment epithelium (nonvisual portion) and a neural portion (visual portion). The pigment epithelium is a sheet of melanin-containing epithelial cells that lies between the choroid and the neural portion of the retina. • melanin in the choroid and the pigment epithelium absorb stray light rays, thereby preventing reflection and scattering of light within the eyeball. This ensures that the image on the retina is sharp and clear. The pigmented layer is continuous over choroid, ciliary body and iris while the nervous layer terminates just before ciliary body. This point is called ora serrata. • The neural portion is thick and composed of four layers

of cells. Beginning from the choroid side, it has a layer of pigmented cells, a layer of receptor cells, a layer of bipolar nerve cells and a layer of ganglion cells. • The receptor cells are called photoreceptors or visual cells. They are of two types: Rod cells and cone cells, named after their shapes. Both have light-sensitive pigments. Specific wavelengths of light breakdown the light-sensitive pigments, and this stimulates the receptor cells to set up nerve impulses.

Major cell types of the retina • The major cell types of retina are described below: – Rod photoreceptor cells are about 120 million, –

– –

–

and specialized for reception in dim light. Cone photoreceptor cells are fewer, about 7 million specialized for sensing bright light and for color vision. There are different cone cell types (each with a different photo-pigment) for each of the three primary colors. These photo-pigments are: chlorolabe - green sensitive, cyanolabe- blue sensitive, Erythrolabe - red sensitive. Horizontal cells interconnect groups of photoreceptor cells. Bipolar cells (at least 4 types, one for the rod cells and one for each type of cone cell) interconnect photoreceptor cells with ganglion cells. Amacrine cells interconnect groups of ganglion cells and bipolar cells. They are unusual neurons

because they have no true axon. They transmit information laterally. – Ganglion cells possess long axons that extend through the nerve fibre layer of the retina and then join together to form the optic nerve. They are the only cell type in the retina possessing long axons (which bundle together to form the nerve fibre layer) and exhibit self-propagated action potentials. – Muller cells are large glial-like cells that extend from the internal limiting membrane (basement membrane) to the external limiting membrane (a region of junction between the muller cells and the photoreceptor cells). The glial cells are very rich in glycogen. Rod cell • The rod cells contain a purplish pigment called visual purple, or rhodopsin. • rhodopsin has two components : scotopsin, a protein moiety and 11-cis retinal, a carotene derivative. When both of these combine they create the conjugated rhodopsin molecule. • They function in dim light and at night. They produce poorly defined images. Cone cell • The cones contain a pigment called visual violet, or iodopsin. • They function in daylight and artificial bright light producing detailed images and colour vision. • The cone cells are not as sensitive as the rod cells and do not respond to dim light. This is why we cannot see colours clearly at night. • The cone cells give colour vision (based on trichromatic theory) because they contain three different pigments, each absorbing light of different wavelengths (red, blue and green colour). • Individual cone cell has one type of pigment. The green and red cones are concentrated in fovea centralis and the blue cones are mostly found outside the fovea and have the highest sensitivity. • The colours, other than these three

primary colours are perceived by the simultaneous stimulation of 2 or all the 3 types of cones. Equal stimulation of all types of cones produces the colour sensation of white. MT BIOLOGY

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2>=24?C