Bio 11 – Exercise 11 – Guttation & Transpiration

H. Guttation

14. Plant 5 viable rice grains previously soaked for 24 hours in a small flower pot with holes at the bottom. Immerse the lower portion of the flower pot in water contained in a shallow dish. Add water to the dish from time to time as the need arises. When the seedlings are 2 to 5 cm long, place the dish with the flower pot on top of a piece of glass or plastic and cover with a bell jar or a transparent, wide-mouthed bottle. Observe the set-up.

14.1 In what part of the leaves are the water droplets observed?

Surface of the blade

14.2 Is the origin of the water droplets and the water seen on the surface of the blade the same? Why?

Yes. Water on the leaves is due to guttation. Water droplets are formed when water vapor, from transpiration,  inside the jar/bottle condense.

14.3 Under natural conditions, at what time of the day is guttation likely to occur?

Dawn and dusk.

I. Transpiration

Potometer Method of Measuring Transpiration

15. Attach a piece of transparent rubber tubing measuring 6-9 mm in internal diameter and about 30 cm long, to a 10-ml burette filled with freshly-boiled and cooled distilled water. Insert a leafy shoot of muntingia or any suitable plant with a stem 6-9 mm in diameter into the other end of the rubber tubing. While this is being done, leave the stopcock of the burette slightly open so that water flows into the tubing. There should be no air bubbles in the tubing after the stem has been fitted in. To hold the rubber tubing in place, tie the end around the stem firmly with a rubber band or a piece of string but not too tightly so as not to constrict the vessels. Clamp the shoot in position. Cover one of the branches with a dry plastic bag.

15.1 Mark and record the level of the water in the burette.

15. 2 After two hours, read the level of water in the burette.

15.3 How much water (mL0 was absorbed by the plant? Express this as transpiration rate (mL/hr)

15.4 How will you prove that transpiration has taken place?

Decrease in water level.

Notes:

Some answers differ for each class/group. You need to do the experiment.

For 15.1, 15.2 and 15.3 : refer to your group’s data.

Bio 11 – Exercise 7 – The Leaf

1.2 Give examples of monocot plants with netted venation

Taro

1.3 Give examples of dicot plants with parallel venation

Nerium oleander

2. Examine prepared slides of monocot, dicot and gymnosperm leaves.

Dicot:

Mesophyll (palisade, spongy) – present

Vascular tissue – enlarged vascular bundle in midrib; bundle cap

Midrib – present

Monocot:

Others – bulliform

Gymnosperm:

Epidermis – upper and lower indistinguishable

Stomata – sunken

Mesophyll – lobed mesophyll cells

Vascular tissue – presence of transfusion tissue

Blade – none

Others – resin duct

4. Examine several modified and specialized leaves and complete Table 7.3

Plant Specimen – Modification – Function

Onion – Bulb – Storage

Poinsettia – Bract – Attract agents of pollination

Pitcher plant – Insect trapping leaves – Nutrition

Peas – Tendril – Support stem

Notes:

Some answers differ for each class/group. You need to do the experiment.

For Table 7.1, specimens given might differ.

For Table 7.2, in some monocot specimens/slides the palisade and spongy might be indistinguishable.

Bio 11 – Exercise 6 – The Stem

1. Study a freshly cut stem of an herbaceous and woody dicot and a monocot plant

1.2 What stem has the most prominent node and internode? Why?

Dicot.

2. Study a vigorously growing coleus plant.

2. 1 Where do new leaves and internodes originate?

Nodes.

3. Study a longitudinal section of a coleus shoot apex under the LPO and observe the following: promeristem, leaf primordium, and primary meristems.

3.1 Tabulate the differences between a shoot tip and root tip.

Root tip has neither a leaf primordia nor axillary buds. Shoot tip has no root cap.

5. Follow the subsequent development of the stem tissues by examining cut sections from the uppermost to the lowermost internodes of coleus.

5.1 In which internode do you find differentiated tissues?

Lower internodes.

6. Study a prepared slide or free-hand section of a monocot stem.

6.2 How do primary tissues of this stem differ from those in coleus?

Monocot vascular bundles are scattered and the ground tissue isn’t partitioned into pith and cortex.

7. Focus on a single vascular bundle of a monocot stem under the LPO.

7.2 What is the pattern of xylem differentiation in the stem?

endarchy

7.3 Tabulate the anatomical differences between a monocot and dicot stem.

Monocot: scattered vascular bundles, no distinction between pith and cortex, no cambium, no secondary thickening.

Dicot: vascular bundles in a ring

8. Examine prepared slides of a woody dicot and pine stems. Study the different tissues under a compound microscope.

8.1 Which tissues constitute the bulk of the stem?

Secondary xylem

8.2 What tissues may have produced the additional cells? How do you account for this?

Vascular cambium. Secondary growth.

8.3 What protective tissue)s) is/are present in your specimen?

Periderm

8.4 What do you call the region with the loosely arranged parenchyma cells? What is its function?

Lenticel for gas diffusion.

8.5 Do you see companion cells and vessel elements in the pine stem?

Yes

8.6 Do you observe resin ducts and secretory cells in your specimens?

Yes

9. Examine cross, radial and tangential sections of a dicot and pine wood.

9.1 Differentiate the arrangement, shape and size of the wood rays in each section of the two specimens.

Cross section: Thin strands

Radial: Box-like

Tangetial: Pod-like

9.2 Tabulate the similarities and differences between a dicot and pine stem

Dicot: No resin canal

10. Examine various modified stems

Plant Specimen – Modification – Function

Water hyacinth – Enlarged – Buoyancy

Potato – Tuber – Storage and Vegetative reproduction

Ginger – Rhizome – Storage

Grass – Stolon – Vegetative reproduction

Cactus – Photosynthetic stem – Photosynthesis

Notes:

Some answers differ for each class/group. You need to do the experiment.

For Table 6.1: Only monocot has leaf sheath and no petiole, only woody dicot has lenticels.

For 7.1:

From subtlerainbow's Multiply album.

Bio 11 – Exercise 5 – The Root

1.1 Classify the specimens according to their types of root systems:

corn (Zea mays L.) – fibrous

mung bean – tap

carrot (Daucus carota L.) – tap

radish (Raphanus sativus L.) – tap

sweet potato – tap

ginger (Zingiber officinale Roscoe) – fibrous

ginger lily (Hedychium coronarium Koen.) – fibrous

taro (Colocasia esculenta (L.) Schott.) or any Araceae – fibrous

onion or any Liliceae or Amaryllidaceae – fibrous

roots or leaves of life plant (Kalanchoe sp.) – fibrous

cuttings of begonia (Begonia sp.) – fibrous

bowstring hemp – fibrous

3. Examine the root system of water hyacinth and note the sequential development of tissues from the tip to the base.

3.1 Where are the youngest roots located? the oldest?

Youngest located at the tip of the root. Oldest located that the base of the root.

3.2 Identify the structure covering the root tip.

Root cap.

4. Examine under the microscope a cross-section of a young monocot root of spiderwort, Commelina benghalensis L. or corn and any dicot root. Locate and note the following primary tissues: epidermis, cortex, stele and pith.

4.1 Do you see the sub-epidermal cells (hypodermis or exodermis) in your specimen? If so, how many layers?

Yes

4.2 In which tissue are the intercellular spaces evident?

Sclerenchyma

4.3 What features make the endodermis morphologically distinct?

Casparian strip

4.4 Determing the number of layers and cell types in the pericycle

Single layer.

4.5 How many protoxylem poles do you see in your specimen?

4.6 Is the pattern of xylem differentiation endarch or exarch?

Exarch.

4.7 Is there a pith in the root core of your specimen?

Yes.

4.8 What tissue gives rise to the lateral root?

Pericycle

4.10 Differentiate a root hair from a branch root

Root hairs are unicellular, epidermal outgrowth which are only temporary. Branch roots are multicellular, endogenous, long lasting and develop from mature roots.

4.11 What region occupies the largest in the root?

Region of elongation

4.12 Tabulate the anatomical differences between monocot and dicot root.

Dicot – tetrarch, shows second growth, phloem between arms of xylem

Monocot – polyarch, vascular bundles arranged in a ring

5. Examine the cross section of a woody dicot root

5.1 Enumerate the cell components of the bark and wood

5.2 Do you find root hairs? Why?

No. Epidermis, where root hair is found, is sloughed off during secondary growth.

5.3 Which tissue usually functions in food storage in an old root? Young root?

Parenchyma

6. Examine the intact (entire) and cross-section of the root or radish, turnip, Pachyrrhizus erosus (L.) Urb. and sweet potato.

6.1 What tissues of these roots are edible?

Secondary tissue produced by cambia and upper part of root

7. Examine other specialized roots and tabulate their functions.

Plant Specimen – Specialized Part – Modification – Function(s)

Radish – Tap – Enlarged, fleshy – Food storage

Carrot – Tap – Enlarged, fleshy – Food storage

Sweet Potato – Tap – Enlarged, fleshy – Food storage

Turnip – Tap – Enlarged, fleshy – Food storage

Mangrove – Fibrous – Aerial root – Support

Pandan – Fibrous – Prop root – Support

8. Examine the cross section of the root of an aerial orchid and study the tissues

8.1 How does its morphology compare with the corn root?

Orchid has velamen which protects the aerial roots.

Notes:

Some answers differ for each class/group. You need to do the experiment.

For 4.1, 4.4 & 4.5, answers depend on your specimen/slide.

Bio 11 – Zoology – Exercise 17 – The Plasma Membrane

How does a living cell like the amoeba remain intact in an aquatic medium?

Contractile vacuole

What similarities can you observe between the milk and the plasma membrane?

Both have regenerative abilities

Does removal of the skin remove any components from the milk preventing further skin formation?

No

How does the organization of a plasma membrane differ from that of the milk?

Milk is made up of proteins. Plasma membrane has a phospholipid bilayer

Does the dye dissolve and diffuse within the protein solution?

Yes

Is there any resemblance to the manner by which particulate material might pass from the outside into the cell’s interior? Explain.

Egg albumen and oil served the same function as the plasma membrane in the cell as a lipid-protein membrane allowing nigrosine to diffuse from a region of high concentration to low.

However, in a living cell, equilibrium is never attained. Explain this phenomenon.

Equilibrium is a steady state achieved when there are no further net movements of moleculesvia any process. Thus, it is only achieved when a cell is dead.

Compare results with those obtained in distilled water.

Distilled water – hemolysis – hypotonic solution

10% NaCl – crenation – hypertonic solution

0.9% NaCl – normal – isotonic solution

Note:

Can expound answer for sixth question by researching on polarity.

Bio 11 – Botany – Exercise 14 – Respiration

A. Release of Heat Energy during Respiration

1. Place germinating mung bean seeds previously soaked in distilled water overnight inside a thermos flask or Dewar bulb and suspend a vial containing 20% solution of potassium hydroxide. Stopper the flask with a cork through which a thermometer is inserted. Similarly prepare another flask as control using boiled seeds previously soaked overnight in 5% formalin.

1.2 Why is potassium hydroxide used?

Used to absorb CO2

B. Release of CO2

2. Fill an Erlenmeyer flask with yeast suspension and two other flasks with saturated barium hydroxide solution. One flask with the barium hydroxide solution is tightly sealed with a rubber stopper while the other two are provided with tight fitting rubber stoppers., each with one hole so as to connect the two flasks with a bent glass tubing. The connection is made alright by sealing the holes with melted candles or vaseline. Observe the results after 24 hours.

2.1 From the set-up, what observations indicate that a reaction has occurred?

White precipitate forms in the flask with barium hydorxide solution that is connected to the flask with yeast.

2.2 Explain your results

CO2 + Ba(OH)2 -> BaCO3 + H2O

C. Necessity of oxygen in Aerobic Respiration

3. Place 10-20 bean seeds, previously soaked in water overnight, on two wads of moist cotton. Wrap each wad with gauze. Suspend one in a corked bottle containing a little water, the other in a similar bottle containing alkaline pyrogalllol solution. keep the bottles in a dark warm place for 48 hours.

3.1 Compare the development of the seeds in each of the bottles. Explain your results.

Alkaline pyrogallol absorbs H2O and O2. Seeds did not germinate in the bottle with alkaline pyrogallol.

D. Factors Affecting the Rate of Anaerobic Respiration (Fermentation)

4. Obtain 3 clean fermentation tubes. Fill the first with 10 ml yeast-glucose suspension, the second with yeast suspension and the third with glucose solution.

4.1 Observe the amount of gas produced at half hour intervals during the experiment. Compare the volume of gas displaced in each fermentation tube.

4.2 Account for differences observed

Yeast uses glucose for fermentation therefore the yeast-glucose suspension produces more gas due to fermentation.

5. Fill 3 fermentation tubes with 10 ml of yeast-glucose solution. Place 1 tube in a water bath of 4 degrees Celsius, the second in a  beaker of ice water and the third at room temperature.

5.2 Explain the differences in the amount of gas produced in the tubes.

Temperature is directly proportional to rate of fermentation.

Note:

Some answers differ for each class/group. You need to do the experiment.

For 1.1, 1.3, Table 14.1 and Table 14.2, you need your group’s data.

Bio 11 – Botany – Exercise 12 – Photosynthesis

A. Chlorophyll as a Factor in Photosynthesis

1. Examine young variegated leaves of coleus or caricature plant, Graptophyllum pictum (L.) Griff, and note the distribution of chlorophyll as indicated by the green area.

2. place the leaf in boiling water for a few minutes. Immerse it in a test tube containing 95% ethyl alcohol. Place the test tube in a warm bath until the pigments are extracted.

2.1 What is the purpose of boiling the leaf?

Burst the chloroplasts

3. Rinse the bleached leaf with water and place in a petri dis. Test for the presence of starch using dilute IKI solution. The presence of starch as shown by a bluish-black to purple coloration indicates the occurence of photosynthesis.

3.2 How does it compare with your sketch in 1.1?

Green ares – photosynthetic

Non-green areas – non-photosynthetic

3.3 Do the non-green areas of the leaf contain starch?

Yes although minimal

3.4 What can you conclude from this experiment?

Only those with chlorophyll can undergo photosynthesis

B. Light as a Factor in Photosynthesis

4. Place a potted coleus or bean plant in the dark. After 48-75 hours, warp one of the leaves with carbon paper or aluminum foil. Expose the plant to light for 6 to 10 hours. Detach the covered and one uncovered leaf of the same size and test for presence of starch as previously done in section A above.

4.1 Why was it necessary to place the plants in the dark for 48-72 hours?

Consume starch reserves in plant

4.2 Why was carbon paper or aluminum foil used?

Prevent light from reaching leaves

4.3 What areas in the covered and uncovered leaves showed a positive reaction to the starch test?

Uncovered areas contained starch

4.4 Can plants carry out photosynthesis under artificial light as well as natural or day light?

Yes

4.5 What can you conclude from this experiment?

Light is needed for photosynthesis

5. Immerse a hydrilla shoot with the cut end of the stem turned up in a clean test tube with tap water. To enrich the CO2 content of the water, add 10-15 drops of 1 percent NaHCO3. place the tube under dim light and count the number of bubbles released at 1 min intervals for 10 mins. Repeat the experiment under bright light.

5.2 How would you correlate the rate of gas evolution with the rate of photosynthesis?

directly proportional

5.3 What is the effect of light intensity on the rate of photosynthesis?

More intense light, higher rate of photosynthesis

C. Carbon Dioxide as a Factor in Photosynthesis

Half-fill each of three test tubes with water that has previously been boiled to remove dissolved air and cooled to room temperature. Add several drops of phenol red to each tube and shake gently. Phenol red is an indicator of p H. Indirectly it is an indicator of CO2 concentration since CO2 in water forms carbonic acid which lowers the pH of water. At low pH, the color of phenol red changes to yellow. With the aid of a soda straw, blow into the water in test tubes 1 and 2 until the solution just shows a change in color from red to yellow. Test tube 3 serves as control.

6.1 Account for any change in color in test tubes 1 and 2

Increase in acidity

7. Using the set-up prepared in section 6, place a shoot of hydrilla in test tube 1. place the 3 tubes in very bright light. After 20 to 30 minutes, note the color changes.

7.1 Which tubes show a change in color? Why?

Color in test tube 1 goes back to red since CO2 is being used for photosynthesis by the hydrilla plant. With less CO2, the water’s acidity goes back to normal level.

D. Separation of Pigments

8. Secure 5 mature leaves of papaya or 10 of hibiscus. Cut the lamina into small pieces and grind in a mortar to secure a deep green liquid extract and collect this into a small test tube. Place two or three drops of the leaf extract at about 1 to 2cm fromthe base of a strip of filter paper the width of which should not touch the sides of the test tubes. Carefully hang the strip on the lower or basal end of a cork with the use of a aper clip or adhesive and lower this into a clean dry test tube to which has been added 3-5ml of a solvent consisting of a mixture of 95 parts petroleum ether and 5 parts of acetone. The part of the strip containing the leaf extract should not be submerged in the solvent. Cork the test tube and observe the separation of the pigments. The development of the chromatogram is stopped when the solvent has travelled about 1 to 2 cm from the top of the strip. The chromatogram is stopped when the solvent has travelled about 1 to 2 cm from the top of the strip. The chromatogram should be observed at frequent intervals because if the separation continues for too long some of the pigments may be superimposed on each other near the top of the strip.

8.1 What colors are indicated in the chromatogramand what are the corresponding pigments they represent?

orange – carotene

yellow – xantophyll

yellow green/ light green – chlorophyll b

green / blue green – chlorophyll a

Note:

Some answers differ for each class/group.

Table 12.1 is based on your group’s data.

For 8.1, it is possible for the chromatogram to not show an orange color.

For “8.2. Which pigment is the least soluble or moves the slowest along the strip? The most soluble and travelling farthest up the strip?” Least soluble would be the color at the bottom and the most soluble would be the one at the topmost.

NaHCO3 was used to supply CO2 in order to drive photosynthesis.

Bio 11 – Botany – Exercise 11 – Plant Water Relations

A. Diffusion

1. Steam for 10 seconds three 0.5 cm cubes of yam roots resting on a wire mesh or strainer. Place this in a mesh or strainer. Place this in a 500-mL beaker 3/4 filled with watr. Prepare a similar set-up using unsteamed cubes. leave both set-ups undisturbed and observe after 30 mintues and at the end of the laboratory period.

1.1 Compare the color intensities in different parts of the beakers at the start of the experiment and after each observation.

Steamed yam roots – red color more intense

1.2 How do the two treatments compare in the rate of spread of the color?

Steamed yam roots – faster

1.3 How would you account for the difference?

Increase in temperature leads to increase in kinetic energy, which leads to an increase in rate of diffusion.

B. Osmosis

3. Put into 8 small test tubes 10 ml of each of the following sucrose solutions: 5 %, 10%, 15%, and 20%. With a scalpel and a pair of forceps carefully peel off uniform small strips of the lower epidermis of leaves o boat of Moses. Examine the strip of the epidermal tissue to verify that it consists of intact cells containing the violet-colored anthocyanin pigments. Drop a strip into each test tube, at about 5-minutes intervals between successive tubes. After a 30-minute immersion of each strip, examine the cells under the microscope, mounting the strip in the same solution in which it has been immersed. Examine one microscope field underLPO and count the number of plasmolyzed cells as indicated by the shrinkage of the cytoplasm.

3.1 What is plasmolysis?

Loss of water from the cell within plants by osmosis

4. Prepare another section of the lower epidermis of boat of Moses. Examine the section under the LPO and select a thinner portion where the stomata are clearly seen, wide open and free from dirt and foreign matter. Focus on one cell with the widest stoma under HPO.

4.4 Explain the effect of sugar solution on the changes in pore size.

Sugar solution is hypotonic. In a hypotonic solution, cells are plasmolyzed. The guard cells sag, inner cell walls come closer and so the stoma closes.

Notes:

Data for Table 11.1 (under Osmosis) is different for each group and class. You’ll need data from your class. The sketch for 3.4 *under Osmosis) is pretty easy – just like how your instructor/ lecture professor drew it. The sketches for 4.1 and 4.3 (under Osmosis) are like how your instructor drew them.

No answer for”4.5 What factors and structures in the guard cells are responsible for the opening and closing of the stomata”. Just expound on answer for 4.4

Bio 11 – Botany – Exercise 2 – The Plant Cell

1. Place a drop of water at the center of the clean slide and place a thin layer of skin peeled off from the inner surface of onion bulb, Allum cepa L. With a dissecting needle, put a cover slip on the specimen by tilting it along one side of the slide so that one edge touches the water. Gently lower the cover slip by withdrawing the needle slowly to prevent air bubbles.

2. Examine one cell under the HPO.

2.1 Is the cell wall uniform in thickness?

Yes

2.2 Can you observe the narrow canals or depressions along the walls? If yes, identify these structures.

Plasmodesmata

2.3 Can you distinguish the protoplast clearly?

No

2.4 Can you see the cytoplasmic strands? No

3. Using filter paper as absorbent, remove the water from the slide by gently pressing one side of the coverslip. Immediately apply a drop of iodo-potassium iodide solution on the opposite side of the coverslip. The solution will slowly replace the water. After a minute or two,wipe off the excess solution

3.1  Do you observe the cytoplasmic strands now?

Yes

3.2 In what part of the cell is the nucleus found?

Cytoplasm

3.3 Do you see more than one nucleolus?

Yes

3.4 What structure(s) is/are between the cytoplasmic strands?

Organelles

4. Take a leaf from the actively growing shoot of a water plant, hydrilla, Hydrilla verticillata (Roxb.) Royle which has been previously exposed to bright light. Make a water mount by placing the upper surface of the leaf next to slide. By moving the  LPO up and down, you can see two layers of cells. Under the HPO observe the streaming protoplast (cyclosis) and note the direction of its movement.

4.1 What structure in the cell contains the green pigment?

Chloroplast

4.2 What is the shape and arrangement of the structure?

In stacks (grana)

4.3 Do they move along with the streaming protoplast?

Yes

6. On a slide, make separate mounts of the skin and pulp of a ripe fruit of tomato, Lycopersicon lycopersicun (L.) Karsten and red pepper, Capsicum frutescens L.

6.1 What structure in the cell contains the pigments?

Chromoplast

7. Make water mounts of both upper and lower surfaces of the leaf of boat of Moses, Rhoeo discolor (Hert.) Hence. One surface contains a vacuole pigment and the other a plastid.

7.1 What types of pigment are found on each surface?

Anthocyanin – vacuole

Chlorophyll – chloroplast

7.2 How do these pigments differ from those of the tomato pulp and hydrilla leaf?

Differ in location and color.

8. Prepare free-hand sections of any 5 of the following specimens representing the different types of crystals: petioles of Begonia sp.; castor oil plant, Ricinus communis L.; the laminae of santan, Ixora sp.; bowstring hemp, Sanseviera zeylanica Roxb.; the midribs of fringed waterplant, Raphidophora merili Engl.; dumbcane, Diefferenbachia sp.; guava, Psidum guajava L.; pigweed, Amaranthus viridus l.; purslane, Portulaca oleracea L.; sweet potato, Opomoea batatas Lam. and stem of Pilea cadieri.

8.1 Identify and sketch the types of crystals observed.

9. Examine prepared slides of a cross-section of any stem or root and fresh surface sections of the lower epidermis of the leaf of the boat of Moses under the LPO and HPO. Locate cells that ar bounded by a single cell wall called the primary cell wall. Locate the middle lamella (the intracellular layer) between the primary walls of adjacent cells.

9.1 What must be the function of the middle lamella?

Cement adjacent walls together

10. Examine fresh sections of green and ripe tomato fruits.

10.1 Compare the appearance of the cells

Cells in the unripe tomato fruit are closer together

10.2 What happens to the middle lamella when fruits ripen

Loss of cell wall components particularly pectin, which makes up the middle lamella

11. Scrape cells from a shell of a coconut, Cocos nucifera L. and stain with 18% alcoholic phloroglucinol-sulfuric acid solution. Examine under the compound microscope. Lignin will turn red with this solution. These cells have both the primary and secondary walls.

11.1 Draw a cell and label parts

a basic sketch of cell from coconut husk (sclereid cells)

Notes:

Some questions are dependent on the specimen and therefore differs for each student. Questions 2.2,. 2.4, 3.1, 3.3 may be answered yes/no dependent on what you observed.

For the types of crystals:

CaC₂O_4:

prismatic

raphides

druses

styloids

CaCO_3:

cystolith

wormlike cystolith

Your instructor will draw a basic sketch of what these crystals look like and that’s pretty much it.

Bio 11- Botany – Exercise 1 – Microscopy

A. The Compound Microscope

5.1 How does the image move in the field of view, in relation to the movement of the slide?

Opposite direction

5.2 From these observations, what can you deduce about the path of light in a body tube?

Light is refracted

6.1 Is the position of the object in the field changed by shifting from LPO to HPO? Why?

Yes

6.2 Is there a decrease or an increase in the area of the specimen seen in the field of vision when changing from LPO to HPO? Why?

Yes. Increase in magnification.

Table 1.1 Parts and Functions of a Compound Microscope

Ocular lens / Eyepiece – lens located at the upper end of body tube to look through

Tube – connects eyepiece to objective lens

Arm – supports tube and connects to base

Stage – platform where the slide is placed

Stage clips – used to hold the slide in place

Revolving nosepiece / turret – holds 2 or more objective lenses and can be rotated to easily change objective power

Objective lenses – used for magnification along with the eyepiece. LPO/HPO/Oil immersion

Dust tube – protects objective power

Course adjustment knob – used to raise or lower objective power

Fine adjustment knob – used for critical focusing

Base – bottom of microscope. Used for support

Diaphragm – controls the amount of light that will enter the microscope

Mirror – used to reflect light from an external light source up through the bottom of the stage

Mirror rack – holds mirror in place

Inclination joint – used to incline the microscope

B. The Dissecting Microscope

8.1 How is the movement of the slide related to that of the image? Why?

Moves in same direction. Single lens.

Note: The next few questions are dependent on the microscope/specimen used and is therefore different for each student. You’ll have to actually do the experiment, sorry.