OM GANESHAYA NAMAH: PREFACE With my rich experience in teaching to Intermediate, Degree and PG students; academic aspect and administration, I found certain abnormalities in advising the students in preparation of Practical Record. It all appears at the time of correction of the record, that the students should be given well training. Even though, the Practical syllabus is same throughout the University, the Records of one college are quite different from another. Why, such discrepancy is occurring? I request the academicians to think over the matter and advice for a better uniformity. In some colleges, I found very bulky records, with the diagrams irrelevant. In my view, we should suggest the students to draw whatever they have seen in the laboratory, either microscopic or macroscopic specimens or slides. What is not observed need not be advised to draw. After all ‘ Record’ means ‘Recording the data and images’, but not that has been not observed in the lab The manuals available in the market are not totally suitable for all the institutions of all the Universities. Besides there are several spelling mistakes in the records of the students from rural background and Telugu medium at their lower level. This has provoked me to have a manual suitable for all the colleges of Osmania University. I am sure that this will be more useful to the students while preparing the records uniformly as well as to the faculty members. Suggestions with respect to additions, deletions, and substitutions (not mutations) are invited to the address / e-mail given below:Dr.Laxman Rao Goje,# 8-5-254/1, Srinilayam, Ground Floor,Syndicate Bank Colony, Old Bowenpally,SECUNDERABAD- 500011e-mail: dr_goje@yahoo.co.in.Mob: 9885192143. AUTHOR ACKNOWLEDGEMENTS. I convey my sincere thanks to Sri Girivardhan Reddy, Sri Bapi Reddy, sri Bhaskar Reddy, the Directors of Sai Chaitanya College, Kompally, Secunderabad, RR District for their inspiration of this work. I am also thankful to Dr. K.Satyaprasad, Dr. Seetha Rama Rao, the Head, Dept, of Botany and Dr. Bhadraiah, Chairman, Board of Studies, Dept. of Botany, Osmania University, Hyderabad for their cooperation in completing the work. . I am indeed indebted to Dr. H. Rajendra Prasad, my backbone, for constant encouragement and critical analysis of the compilation of work. My thanks are also due to the other faculty members of the Sai Chaitanya College, Kompally, Secunderabad and Dept. of Botany, Osmania University, Hyderabad. Finally, I am also thankful to members of my family like my wife, Mrs.Indira; Chaitanya, and Vignyan, sons and Mrs. Sridevi, daughter-in-law, Master Tanishq, grand son for their cooperation in completion of this work. Dr. L. R. Goje.OSMANIA UNIVERSITY, HYDERABADB.Sc., III Year, Paper-III.Model Practical Question Paper.Time: 3 hr Max. Marks:50Q.I. Make the cytological preparation of the given material and describe any 2 stages with diagrams. ( Procedure-3, Slide preparation-6, Description & Importance-3, diagrams-3)….15QII. Solve the following TWO genetic problems. (5x2)………………………………………...10 ( One from Mono- or Dihybrid, complete or Incomplete dominance OR interactions; One from Linkage & Mapping)QIII. Conduct the given Ecology experiment. ……………………………………05 ( Either Soil Texture & pH OR Estimation of CO3-2 , HCO3- )QIV.Spots: 5 spots x each 2. ………………………10In Cell Biology & Genetics; 2 spots from instruments of Horticulture; andspots from Ecology.QV. Record & Viva/ Field trip 5+5…………………….10 CELL BIOLOGYFixation (histology)In the fields of histology, pathology, and cell biology, fixation is a chemical process by which biological tissues are preserved from decay, either through autolysis or putrefaction. Fixation terminates any ongoing biochemical reactions, and may also increase the mechanical strength or stability of the treated tissues.Purpose of fixationThe purpose of fixation is to preserve a sample of biological material (tissue or cells) as close to its natural state as possible in the process of preparing tissue for examination. To achieve this several conditions must usually be met.First, a fixative usually acts to disable intrinsic biomolecules – particularly proteolytic enzymes – which would otherwise digest or damage the sample.Second, a fixative will typically protect a sample from extrinsic damage. Fixatives are toxic to most common microorganisms (bacteria in particular) which might exist in a tissue sample or which might otherwise colonise the fixed tissue. In addition, many fixatives will chemically alter the fixed material to make it less palatable (either indigestible or toxic) to opportunistic microorganisms.Finally, fixatives often alter the cells or tissues on a molecular level to increase their mechanical strength or stability. This increased strength and rigidity can help preserve the morphology (shape and structure) of the sample as it is processed for further analysis.Even the most careful fixation does alter the sample and introduce artifacts that can interfere with interpretation of cellular ultrastructure. A prominent example is the bacterial "mesosome", which was thought to be an organelle in gram-positive bacteria in the 1970s, but was later shown by new techniques developed for electron microscopy to be simply an artifact of chemical fixation. Standardization of fixation and other tissue processing procedures takes this introduction of artifacts into account, by establishing what procedures introduce which kinds of artifacts. Researchers who know what types of artifacts to expect with each tissue type and processing technique can accurately interpret sections with artifacts, or choose techniques that minimize artifacts in areas of interest.Fixation processFixation is usually the first stage in a multistep process to prepare a sample of biological material for microscopy or other analysis. Therefore, the choice of fixative and fixation protocol may depend on the additional processing steps and final analyses that are planned. For example, immunohistochemistry utilizes antibodies which bind to a specific protein target. Prolonged fixation can chemically mask these targets and prevent antibody binding. In these cases, a 'quick fix' method using cold formalin for around 24 hours is typically used.Types of fixationHeat fixation: After a smear has been allowed to dry at room temperature, the slide is gripped by tongs or a clothespin and passed through the flame of a Bunsen burner several times to heat-kill and adhere the organism to the slide.Chemical FixationProcess whereby structures are preserved in a state (both chemically and structurally) as close to living tissue as possible. This requires a chemical fixative which can stabilise the proteins, nucleic acids and mucosubstances of the tissue by making them insoluble.Types of Chemical FixativesCrosslinking fixatives - AldehydesCrosslinking fixatives act by creating covalent chemical bonds between proteins in tissue. This anchors soluble proteins to the cytoskeleton, and lends additional rigidity to the tissue.By far the most commonly used fixative in histology is formaldehyde. It is usually used as a 10% Neutral Buffered Formalin (NBF). For quick clarification, Formaldehyde is a gas. Formalin is formaldehyde gas dissolved in water. Paraformaldehyde is a polymerised form of formaldehyde, usually obtained as a fine white powder, which depolymerises back to formalin when heated. Formaldehyde fixes tissue by cross-linking the proteins, primarily the residues of the basic amino acid lysine. Its effects are reversible by excess water and it avoids formalin pigmentation. Other benefits include: Long term storage and good tissue penetration. It is particularly good for immunohistochemistry techniques. Also the formaldehyde vapour can be used as a fixatives for cell smears.Another popular aldehyde for fixation is glutaraldehyde. It operates in a similar way to formaldehyde by causing deformation of the alpha-helix structures in proteins. As a somewhat larger molecule, glutaraldehyde may not penetrate thicker tissue specimens as effectively as formaldehyde. Thus small blocks of tissue are required. On the other hand, glutaraldehyde may offer a more rigid or tightly linked fixed product—its greater length and two aldehyde groups allow it to 'bridge' and link more distant pairs of protein molecules. It causes rapid and irreversible changes, fixes quickly, is good for electron microscopy, fixes well at 4oC, and gives best overall cytoplasmic and nuclear detail. However it is not ideal for immunohistochemistry staining.Some fixation protocols call for a combination of formaldehyde and glutaraldehyde, so that their respective strengths complement one another.These crosslinking fixatives – especially formaldehyde – tend to preserve the secondary structure of proteins and may protect significant amounts of tertiary structure as well.Precipitating fixatives - AlcoholsPrecipitating (or denaturing) fixatives act by reducing the solubility of protein molecules and (often) by disrupting the hydrophobic interactions which give many proteins their tertiary structure. The precipitation and aggregation of proteins is a very different process from the crosslinking which occurs with the aldehyde fixatives.The most common precipitating fixatives are ethanol and methanol. They are commonly used to fix frozen sections and smears. Acetone is also used and has been shown to produce better histological preservation than frozen sections when employed in the Acetone Methylbenzoate Xylene (AMEX) technique.The protein denaturants - methanol, ethanol and acetone - are rarely used alone for fixing blocks unless studying nucleic acids.Acetic acid is a denaturant that is sometimes used in combination with the other precipitating fixatives. The alcohols, by themselves, are known to cause considerable shrinkage and hardening of tissue during fixation while acetic acid alone is associated with tissue swelling; combining the two may result in better preservation of tissue morphology.Oxidising agentsThe oxidising fixatives can react with various side chains of proteins and other biomolecules, allowing the formation of crosslinks which stabilize tissue structure. However they cause extensive denaturation despite preserving fine cell structure and are used mainly as secondary fixatives.Osmium tetroxide is often used as a secondary fixative when samples are prepared for electron microscopy. (It is not used for light microscopy as it penetrates thick sections of tissue very poorly.)Potassium dichromate, chromic acid, and potassium permanganate all find use in certain specific histological preparations.MercurialsMercurials such as B-5 and Zenker's have an unknown mechanism which increases brightness of staining along with giving excellent nuclear detail. Despite being fast, mercurials penetrate poorly and produce tissue shrinkage. Their best application is for fixation of hematopoietic and reticuloendothelial tissues. Also note that since they contain mercury care must be taken with disposal.PicratesPicrates penetrate tissue well to react with histones and basic proteins in order to form crystalline picrates with amino acids and precipitate all proteins. It is a good fixative for connective tissue, preserves glycogen well and extracts lipids in order to give superior results to formaldehyde in immunostaining of biogenic and polypeptide hormones However it causes a loss of basophilia unless the specimen is thoroughly washed following fixation.HOPE FixativeHepes-glutamic acid buffer-mediated organic solvent protection effect (HOPE) gives formalin-like morphology, excellent preservation of protein antigens for immunohistochemistry and enzyme histochemistry, good RNA and DNA yields and absence of crosslinking proteins.Frozen SectionsSmall pieces of tissue (5X5X3mm) are placed in a cryoprotective embedding medium - OCT, TBS or Cryogel - then snap frozen in isopentane cooled by liquid nitrogen. Tissue is then sectioned in a freezing microtome or cryostat. Sections are then fixed in one of the following fixatives: Absolute acetone for 10-15 minutes, 95% ethanol for 10-15 minutes or Absolute acetone 10minutes followed by 95% ethanol 10minutes1. MITOSIS ( Slide preparation and identification of stages).AIM: Preparation and study of slides for mitosis using the ‘squash techniq- ue’ from the root tips.Acetocarmine method( WARMKE’S METHOD,1935)Requirement: Fixed onion root tips, acetocarmine, 45% acetic acid, slid-es, cover slips, filter paper, spirit lamp.METHOD: 1. Cut the distal end of the root tips in pretreatment solution (95% etanol & conc.HCl in 1:1) for 5 minutes and transfer again in Carnoy’sFluid for 5 min.2. Stain the tips in acetocarmine and put a cover slip.3. Squash it by gently heating and pressing, with a few folds of filter paper.4. Study the different mitotic stages and draw the diagrams.PREPLAN: For getting fixation of onion root tips, one has to plan a week before. The onions have to be kept on a beaker of water, using three toothpicks , so that the base of the bulb touches the water, as shown in the diagram. FIXATION: The fixing fluid is Carnoy’s-I fluid,( as given- Absolute ehanol=30ml + glacial acetic acid =10ml). The root tips have to be kept only for 15 minutes, then transferred to 60% ehanol preservative ( keep at 40 C).MITOSIS:1. It is a part of Somatic Cell Division.2. It include, the division of nucleus (=Karyokinesis) and the division of cytoplasm (=Cytokinesis).3. Strasburger (1875), a German botanist, was the first to work out the details of mitosis.4. The most favourable material is the apices of onion root tips.5. The karyokinesis includes four important stages, viz.,Prophase, Metaphase, Anapha- se and Telophase.PROPHASE:6. It is the longest stage in Mitosis.7. The nucleus at early stage contain delicate and coiled thread-like structures called Chromonemata.8. The chromonemata starts condensation, and transform into chromosomes, each contain two chromatids coiled around each other. The constriction-like centromeres appear in later stages.9. The nucleolus start decreasing in size and ultimately disappear at the end of prophase. 10. The nuclear membrane also disrupts and disappear. METAPHASE:11. The nuclear spindle appear cotaining bipolar fine and delicate fibrils.12. The chromosomes become shorter and thicker.13. They move to the equatorial plane of the spindle, in the systematic form.14. The maximum condensation is seen in this stage and chromosomes are best seen.15. At the end of the metaphase, the chromatid pair of each chromosome split, containing their centromere repel each other. METAPHASE ANAPHASE ANAPHASE:16. As a result of repulsion, the pair of chromatids in each chromosome start moving towards the two opposite poles. This is due to the contraction of chromosome spi- ndle fibres.17. This is the shortest phase and completed as the two sets of the chromatids reach to the opposite poles.TELOPHASE: 18. The chromatids are now regarded as chromosomes.19. Disappearance of polar caps of the spindle, along with spindle body.20. The activities are reverse to that of prophase.21. The elongation of chromosomes takes place.22. The nuclear membrane forms around each group of chromosomes.23. The nucleolus reappears, along with nucleoplasm.24. The newly formed nuclei enlarges.CYTOKINESIS:25. Phragmoplast forms between the two newly formed nuclei, by the fusion of phra- Gmosomes and pieces of endoplasmic reticulum.26. The phragmoplast widens centrifugally upto the cell wall.27. It develops into middle lamella and later on cellulose material is deposited.ONION BUD MEIOSISAim: studying themeiotic stages in onion flower buds (Allium Canadense), which flowers the last of May. Material: The buds ( which are generally partially replaced by small bulbs, are enclosed in a spathe or sheath which entirely conceals them. Buds two to three millimeters in length should be selected), slides, cover glasses, acetic acid, ethyl alcohol, acetocarmine, glycerine, DPX mountant, blotting papers. Procedure: The flower buds are opened so as to expose the anthers. The latter should now be removed to a slide, and carefully crushed in a drop of dilute acetic acid (one-half acid to one-half distilled water). This at once fixes the nuclei, and by examining with a low power, we can determine at once whether or not we have the right stages. The spore mother cells are recognizable by their thick transparent walls, and if the desired dividing stages are present, a drop of staining fluid should be added and allowed to act for about a minute. The preparation being covered with a cover glass. After the stain is sufficiently deep, it should be carefully withdrawn with blotting paper, and pure water run under the cover glass.If the preparation is to be kept permanently, the acid must all be washed out, and dilute glycerine run under the cover glass. The preparation should then be sealed with Canada balsam or some other cement, but previously all trace of glycerine must be removed from the slide and upper surface of the cover glass. It is generally best to gently wipe the edge of the cover glass with a small brush moistened with alcohol before applying the cement.I Fig: A, pollen mother cell of the wild onion. n, nucleus. B–F, early stages in the division of the nucleus. par.- Nucleolus.II Fig:Later stages of nuclear divisions in the pollen mother cell of wild onion, × 350. All the figures are seen from the side, except B ii, which is viewed from the pole.a-r: Stages of meiosis from squash preparations of pollen mother cells of Allium triquetrum, stained in Feulgens. a, leptotone; b, pachytene; c, diplotene; d, diakinesis of prophase I. Notice the characteristic “beaded” appearance of the synapsed chromosomes in b; the desynapsing chromosomes of c with chiasmata; the 9 compact bivalents in d. In e, metaphase I, the chromatin adjacent to the centromeres (arrows) is markedly drawn out from the chiasmata. f-h, anaphase I. Notice the different forms cf bivalents which can be interpreted according to the number and positions of the chiasmata in each. In g, the chiasmata have been lost (see Fig. 7) except that of the right hand bivalent (upper arrow): also note the widely separated chromatids of each chromosome (lower arrow). i, telephase I. In j, interphase, the new cell wall separating the nuclei is clear; also note the original parent cell wall. k, prophase II; l, metaphase II; m, one diad at metaphase II, the other anaphase II; n-p, anaphase II; q, telophase II with beginnings of cell plate formation; r, tetrad of pollen grains before liberation from the parent cell wall. a-g have been squashed during preparation to show the relevant features more clearly. Nucleoli are not visible as they do not stain with Feulgens.Prophase IHomologous chromosomes pair and form synapsis, a step unique to meiosis. The paired chromosomes are called bivalents or tetrads, which have two chromosomes and four chromatids, with one chromosome coming from each parent. At this stage, non-sister chromatids may cross-over at points called chiasmata.Leptotene1) The first stage of prophase I is the leptotene stage, also known as leptonema, from Greek words meaning "thin threads.2) During this stage, individual chromosomes begin to condense into long strands within the nucleus. However the two sister chromatids are still so tightly bound that they are indistinguishable from one another. 3) The chromosomes in the leptotene stage show a specific arrangement where the telomeres are oriented towards the nuclear membrane. Hence this stage is called, "bouquet stage".ZygoteneThe zygotene stage, also known as zygonema, from Greek words meaning "paired threads,"[1] occurs as the chromosomes approximately line up with each other into homologous chromosomes. The combined homologous chromosomes are said to be bivalent. They may also be referred to as a tetrad, a reference to the four sister chromatids. The two homologous chromosomes become "zipped" together, forming the synaptonemal complex, in a process known as synapsis.PachyteneThe pachytene stage, also known as pachynema, from Greek words meaning "thick threads,"[1] contains the following chromosomal crossover. Nonsister chromatids of homologous chromosomes randomly exchange segments of genetic information over regions of homology. (Sex chromosomes, however, are not identical, and only exchange information over a small region of homology.) Exchange takes place at sites where recombination nodules or chiasmata (singular: chiasma) have formed. The exchange of information between the non-sister chromatids results in a recombination of information; each chromosome has the complete set of information it had before, and there are no gaps formed as a result of the process. Because the chromosomes cannot be distinguished in the synaptonemal complex, the actual act of crossing over is not perceivable through the microscope.DiploteneDuring the diplotene stage, also known as diplonema, from Greek words meaning "two threads,"[1] the synaptonemal complex degrades and homologous chromosomes separate from one another a little. The chromosomes themselves uncoil a bit, allowing some transcription of DNA. However, the homologous chromosomes of each bivalent remain tightly bound at chiasmata, the regions where crossing-over occurred.In fetal oogenesis all developing oocytes develop to this stage and stop before birth. This suspended state is referred to as the dictyotene stage and remains so until puberty. In males, only spermatogonia exist until meiosis begins at puberty.DiakinesisChromosomes condense further during the diakinesis stage, from Greek words meaning "moving through."[1] This is the first point in meiosis where the four parts of the tetrads are actually visible. Sites of crossing over entangle together, effectively overlapping, making chiasmata clearly visible. Other than this observation, the rest of the stage closely resembles prometaphase of mitosis; the nucleoli disappear, the nuclear membrane disintegrates into vesicles, and the meiotic spindle begins to form.Synchronous processesDuring these stages, centrosomes, each containing a pair of centrioles are migrating to the two poles of the cell. These centrosomes, which were duplicated during S-phase, function as microtubule organizing centers nucleating microtubules, essentially cellular ropes and poles, during crossing over. They invade the nuclear membrane after it disintegrates, attaching to the chromosomes at the kinetochore. The kinetochore functions as a motor, pulling the chromosome along the attached microtubule toward the originating centriole, like a train on a track. There are four kinetochores on each tetrad, but the pair of kinetochores on each sister chromatid fuses and functions as a unit during meiosis I. [2][3]Microtubules that attach to the kinetochores are known as kinetochore microtubules. Other microtubules will interact with microtubules from the opposite centriole. These are also nonkinetochore microtubules. KARYOTYPE STUDYGENETICS PROBLEMSMONOHYBRID CROSS:1) In Garden pea, round seed shape (R) is dominant over wrinkled (r). If homozygous rou-nd seeded plant is crossed with wrinkled seeded plant, what is the shape of seeds in F1?If F1 plant is allowed for self pollination, what are the phenotypic and genotypic ratios of F2 ? ROUND X WRINKLED RR r r(G) ® (r) F1 Rr …. Round (heterozygous) ………..Answer F1 Round Rr X F1 Round Rr (G) ® (r) ® (r) F2 RR RrR rr rF2 Genotypic ratio = RR1 : R r 2 : r r 1F2 Phenotypic ratio = Round(RR,Rr) 3 : Wrinkled (rr) 1………….Answer.--------------------------------------------------------------------------------------------------------------------------2) In summer squash, the gene for white colour of the fruit (W) is dominant over the gene foryellow colour (w). What is the phenotype of the offsprings from the following crosses:i) Ww x Ww ii) WW x ww iii) Ww x ww iv) Ww x WW .(i) Parents Ww White x Ww White Gametes (W) (w) (W) (w) Offspring W W W w W w w w Genotypic ratio: W W1 : W w2 : w w 1 Phenotypic ratio: White 3(WW,Ww) : Yellow 1(ww)…………….Answer.(ii) Parents W W White X w w Yellow Gametes (W) (w) Offspring All W w White only …………………………..Answer.(iii) Parents W w White x w w Yellow Gametes (W) (w) (w) Offspring W w w w Ratio White (W w) 1: Yellow (w w) 1………………………Answer.(iv) Parents W w White x W W White Gametes (W) (w) (W) Offspring W W W w Ratio All White ( Half homozygous white WW, half heterozygous white Ww)3) In snapdragons, the red colour of the flower (R) is incompletely dominant over white colour (r). The heterozygous flowers are pink in colour. What are the phenotypes of parents and offsprings obtained from the following crosses.i) Rr x RR ii) RR x rr iii) Rr x Rr iv) Rr x rr( An example of ‘Monohybrid Incomplete Dominance’)(i) Parents Rr Pink x RR Red Gametes ® (r) ® Offspring RR Rr Ratio Red (RR) 1 : Pink (Rr) 1 . . . . . . . . . . . . . . . . . . . . .Answer(i).(ii) Parents RR Red x r r white Gametes ® (r) Offspring R rAll are pink . . . . . . . . . . . . . . . . . . . . . . . .Answer(ii).(iii) Parents Rr pink x Rr pink Gametes ® (r) ® (r) Offspring RR Rr Rr r r Ratio Red (RR)1: Pink (Rr) 2: White(r r) 1. . . . . . . . . . . . .Answer(iii).(iv) Parents Rr pink x r r white Gametes ® (r) (r) Offspring Rr r r Ratio Pink (Rr) 1 : white (r r) 1. . . . . . . . . . . . . . . . . . . . . . Answer(iv)--------------------------------------------------------------------------------------------------------------------------DIHYBRID CROSS: ( Complete dominance )4) In Garden peas, yellow seed colour (Y) is dominant over green colour (y). Round seed shape (R) is dominant over wrinkled seed (r) . The two seed characters assort Independently. If a green wrinkled seeded plant is crossed with pure yellow round seeded plant, what are the phenotype and genotype of F1 off springs ? If F1 offsprings are allowed for self pollination what is the phenotype ratio of the F2 offsprings.Parents- Yellow, Round YY RR x Green, Wrinkled yy r rGametes- (YR) (yr)F1 Yy Rr - All are Yellow, Round. . . . . . .AnswerF1 were allowed for self pollination- ♂ F1 Yellow,Round YyRr x F1 Yellow, Round YyRr♀Because both parents are heterozygous for 2 gene pairs, 4 types of gametes are formed from each parent as ; YR, Yr, yR, and yr. For combination checker boardis necessary:Female/Male YR Y r yR yr YR YYRR 1 YYRr 2 YyRR 3 YyRr 4 Yr YYRr 2 YYrr 5 YyRr 4 Yyrr 6 yR YyRR 3 YyRr 4 yyRR 7 yyRr 8 yr YyRr 4 Yyrr 6 yyRr 8 yyrr 9F2 offspring analysis:____________________________________________________________________S.No GENOTYPE Number PHENOTYPE Number_______________________________________________________________________ 1 YYRR 1 Yellow, Round 2 YYRr 2 -do-9 - Parental 3 YyRR 2 -do- 4 YyRr 4 -do- 5 YYrr 1 Yellow, Wrinkled. 6 Yyrr 2 -do-3 - Recombinant 7 yyRR 1 Green, Round 8 yyRr 2 -do- 3 - Recombinant 9 yyrr 1 Green, wrinkled .... 1 – ParentalF2 Phenotypic Ratio – Yellow,Round 9: Yellow,Wrinkled 3: Green,Round 3: Green, Wrinkled 1. ( 9:3:3:1)-.-.-.-.-.-.-.-.-.-.-.-.-.-.-Answer----------------------------------------------------------------------------------------------------------------------5. What gametes will be formed by the pea plants involved in the following crosses ? Dete-rmine the phenotypic ratio of the offspring from each cross.i) Yy Rr x yy r r , ii) YyRr x yy RR , iii) YyRr x YyRR , iv) YyRR x yyrr , v) YyRr x YYRrvi) YYRR x YYRr.i) Parents YyRr yellow,round x yy rr green, wrinkled Gametes (YR),(Yr),(yR),(yr) (yr) . . . . . . . . . . . . . . . . . . . . . . . .Answer. Offspring YyRr Yyrr yyRryyrr Phenotypes –yellow,round1: yellow,wrinkled1: green,round1: green,wrinkled1…Ans.ii) Parents YyRr yellow,round x yyRR green, round gametes- (YR), (Yr), (yR), (yr) (yR). . . . . . . . . . . . . . . . . . . . . . . . . Answer. Offspring- YyRR YyRr yyRR yyRrPhenotypes- Yellow,round 2: Green,round 2 i.e., 1: 1. . . . . . . . . . . .Answeriii) Parents- YyRr yellow,round x YyRR yellow,round Gametes- (YR), (Yr), (yR), (yr) (YR), (yR). . . . . . . . . . . . . . . . . . Answer. Offspring are formed of 8 genotypes, as given in the following checker board. Female/Male YR Y r yR yr YR YYRR 1 YYRr 2 YyRR 3 YyRr 4 yR YyRR 3 YyRr 4 yyRR 5 yyRr 6Analysis- Sl.No Genotype Number Phenotype Number 1 YYRR 1 Yellow,Round 2 YYRr 1 -do- 6 3 YyRR 2 -do- 4 YyRr 2 -do- 5 yyRR 1 Green,Round2 6 yyRr 1 -do-Phenotypic ratio- Yellow,Round 3 : Green,Round 1 . . . . . . . . . . . . Answer.iv) Parents YyRR yellow,round x yyrr Green, wrinkled Gametes (YR) (yR) (yr)Offspring ratio- YyRr yellow,round 1: yyRr green,round 1. . . . . . . . . .Answer.v) Parents- YyRr Yellow,Round x YYRr Yellow,Round Gametes- (YR), (Yr), (yR), (yr) (YR), (Yr)Female/Male YR Y r yR yr YR YYRR 1 YYRr 2 YyRR 3 YyRr 4 Yr YYRr 2 YYrr 5 YyRr 4 Yyrr 6Analysis- S.No. Genotypes Number Phenotypes Number 1 YYRR 1 Yellow,Round 2 YYRr 2 -do- 3 YyRR 1 -do-6 4 YyRr 2 -do- 5 YYrr 1 Yellow, Wrinkled 6 Yyrr 1 -do- 2Phenotypic ratio- Yellow,Round 3 : Yellow,Wrinkled 1. . . . . . . . . . . . . . Answer.vi) Parents YYRR yellow,round x YYRr yellow,round Gametes (YR) (YR) , (Yr) Offspring ratio- YYRR 1: YYRr 1, all are Yellow,Round. . . . . . . . . . . Answer.6. When a cross was made between two ‘Round shaped’, ‘Pendant fruited’ chilli Plants, the offsprings obtained fall in the following number;Round,Pendant-221; Round,Erect-79; Long,Pendant-73; Long,Erect-24;Find out the genotypes of the parents. Round(R) and Pendant(P) fruit character aredominant over Long(r) and Erect(p) fruits.The offsprings are in the ratio of 221: 79: 73: 24 , i.e., 9:3:3:1 (fractions rounded off)The ratio is obtained only , when both the parents are ‘double heterozygous’ (as AaBb)Hence, the genotype of both the parents is – Rr Pp . . . . . . . . . . . .Answer.If a cross is made as in problem number-4, the ratio 9:3:3:1 is obtained( Note:Student is advised to complete by giving the checker board as in problem -4).(It is an example of a cross between diuble heterozygous F1 parents)--------------------------------------------------------------------------------------------------------------------------7. In garden pea, the yellow colour and round shape of the seed are dominant over greencolour and wrinkled shape. When a yellow round seeded plant is crossed with green wrin-kled seeded plant, the offspring are obtained in the following ratio; yellow,round-13: yellow,wrinkled-14: green,round-14: green,wrinkled-15. What are the genotypes of the parents( Dihybrid test cross)It is an example of ‘Two Point Test Cross’ (2PTC), and the offsprings fall in 1:1:1:1This ratio is obtained, when one parent is double heterozygous and the other parentIs double recessive ( as AaBbxaabb),Hence the genotypes of the parents is YyRr Yellow,Round x yyrr Green,Wrinkled The gametes (YR), (Yr), (yR), (yr) (yr) Offsprings YyRr Yyrr yyRr yyrr Phenotypes- Yellow,Round 1: Yellow,Wrinkled 1: Green,Round 1: Green,Wrinkled 1So the genotypes of parents is YyRr and yyrr . . . . . . . . . . . . . . . .Answer.---------------------------------------------------------------------------------------------------------------------------8. In Antirrhimum Red flowers (R) are incompletely dominant over white flowers (r), hetero-zygotes produce Pink flowers(Rr). Broad leaves (B) are incompletely dominant over narrowleaves (b), heterozygotes produce medium leaves (Bb). If a red flower, broad leaved plant iscrossed with white flower , narrow leaved plant, what is the phenotypic F1 offsprings ? Givephenotypes and genotypic ratio of the F2 offsprings, when F1 plants are crossed .Parents RRBB Red,Broad x rrbb White, NarrowGametes (RB) (rb)F1 offspring RrBb - Pink flower,medium leaved. . . . . . . . .AnswerCross between F1- - - RrBb pink,medium x RrBb pink, mediumEach parent produces 4 types of gametes, because both are double heterozygous asRB,Rb,rB and rb. For mating again 4x4 checker board is to be drawn.Female/Male RB Rb rB rb RB RRBB 1 RRBb 2 RrBB 3 RrBb 4 Rb RRBb 2 RRbb 5 RrBb 4 Rrbb 6 rB RrBB 3 RrBb 4 rrBB 7 rrBb 8 rb RrBb 4 Rrbb 6 rrBb 8 rrbb 9Analysis; S.No. Genotype Number Phenotype Number 1 RRBB 1 Red,Broad 1 2 RRBb 2 Red,Medium 2 3 RrBB 2 Pink,Broad 2 4 RrBb 4 Pink,Medium 4 5 RRbb 1 Red,Narrow 1 6 Rrbb 2 Pink,Narrow 2 7 rrBB 1 White,Broad 1 8 rrBb 2 White,Medium 2 9 rrbb 1 White,Narrow 1The phenotypic ratio is same like genotypic ratio, i.e., 1:2:2:4:1:2:1:2:1. . . . . . . Answer.---------------------------------------------------------------------------------------------------------------------------GENE INTERACTIONS:9. A pure ‘Rose’ combed chicken is mated with a pure ‘Pea’ combed chicken. All the F1 are‘Walnuts’. Cross F1 walnut with Rose and Pea separately and show the phenotypes andgenotypes.Two pairs of alleles are acting on the character of comb. A=Rose, B=Pea, A_B_=WalnutAabb= Single comb.Parents- Rose AAbb x Pea aaBBGametes - (Ab) (aB)F1 offspring AaBb WalnutsA male F1 walnut is crossed with a female F1 walnut. ♂ F1 Walnut AaBb x F1 walnut AaBb ♀Each parent is double heterozygous, and produces 4 types of gametes. Checker Board has to be drawn.Female/Male AB Ab aB ab AB AABB 1 AABb 2 AaBB 3 AaBb 4 Ab AABb 2 AAbb 5 AaBb 4 Aabb 6 aB AaBB 3 AaBb 4 aaBB 7 aaBb 8 ab AaBb 4 Aabb 6 aaBb 8 aabb 9F2 Analysis is given belowSl.No. Genotypes Number Phenotypes Number_______________________________________________________ 1 AABB 1 Walnut 2 AABb 2 -do-9 3 AaBB 3 -do- 4 AaBb 4 -do- 5 AAbb 1 Rose 6 Aabb 2 -do-3 7 aaBB 1 Pea 8 aaBb 2 -do-3 9 aabb 1 Single……… 1F2 Phenotypic ratio= Walnut 9: Rose 3 : Pea 3: Single 1. . . . . . . . . .Answer.( An example of Collaboration)--------------------------------------------------------------------------------------------------------------------------10) In sweet pea, genes C & P are necessary for coloured flowers. The absence of either or both of these genes, the flowers are white. What will be the ratio of offsprings of thefollowing crosses. a) Ccpp x ccPp, b) CcPp x Ccpp, c) CcPp x CcPp.a) Parents Ccpp White x ccPp White Gametes (Cp) (cp) (cP) (cp) Offspring CcPp Ccpp ccPp ccpp Ratio- Coloured 1(CcPp) : White 3 (Ccpp,ccPp, ccpp). . . . . . . . . . Answer(a)b) Parents CcPp Coloured x Ccpp WhiteGametes-(CP), (Cp), (cP), (cp) (Cp), (cp) Checker board has to be drawn: Female/Male CP Cp cP cp Cp CCPp 1 CCpp 3 CcPp 2 Ccpp 4 cp CcPp 2 Ccpp 4 ccPp 5 ccpp 6Analysis- Sl.No Genotype Number Phenotype Number 1 CCPp 1 Coloured 3 2 CcPp 2 -do- 3 CCpp 1 White 4 Ccpp 2 -do- 5 5 ccPp 1 -do- 6 ccpp 1 -do- Phenotypic ratio = Coloured 3 : White 5 . . . . . . . . . . . . . . . . . . . .Answer(b)c) Parents CcPp Coloured x CcPp coloured Each parent is double heterozygous, hence form 4 types of gametes Female/Male CP Cp cP cp CP CCPP 1 CCPp 2 CcPP 3 CcPp 4 Cp CCPp 2 CCpp 5 CcPp 4 Ccpp 6 cP CcPP 3 CcPp 4 ccPP 7 ccPp 8 cp CcPp 4 Ccpp 6 ccPp 8 ccpp 9Coloured analysis Sl.No Genotype Number Phenotype Number 1 CCPP 1 Coloured 2 CCPp 2 -do- 3 CcPP 2 -do- 9 4 CcPp 4 -do- 5 CCpp 1 White 6 Ccpp 2 -do- 7 ccPP 1 -do-7 8 ccPp 2 -do- 9 ccpp 1 -do- Phenotypic ratio= Coloured 9 : White 7. . . . . . . . . . . . . . . . . . . .Answer(c)11. In mice, black colour of hair is determined by a dominant gene C. Agouti is a wild char-acter which is dependant on dominant gene A. This wild character is expressed when everit interacts with coloured gene. Albino mice are with recessive genes. Find out the ratios of F1 and F2 offsprings resulting from a cross between black and albino mice.(The phenotypes are- A-C-=Agouti, aacc=Albino, aaC-=Black)Parents- Black aaCC x Albino aaccGametes- (aC) (ac)F1 offspring aaCc Black. . . . . . . . . . . . . . . . . . . . . . . AnswerCross between F1 offspring ♂ F1 Black aaCc x F1 Black aaCc ♀Gametes (aC) (ac) (aC) (ac)F2 offspring aaCC aaCc aaCc aaccF2 Ratio Black 3 ( aaCC1,aaCc2) : Albino 1 (aacc) . . . . . . . . . .Answer.13. In Shepherd purse, triangular fruits are dependant on either one or two dominant genes.Top shaped fruits are recessive. A cross was made between two triangular fruited plants.What will be the fruit shape of offspring ? (Triangular-AABB,AABb,AaBB,AaBb, Aabb,aaBbParents- Triangular AAbb x Triangular aaBB aaBB,AAbb; Top-aabb)Gametes- (Ab) (aB)Offspring- AaBb All Triangular. . . . . . . . . . . AnswerLINKAGEAND CHROMOSOME MAPPING14. A cross was made between doubly heterozygous purple flower, long pollen variety with red flower, round pollen variety produced the following offspring: Purple,long-40; Purple,round-10; Red,Long-10; Red,Round-40;a) Show the parental cross on the chromosome,b) Calculate the percentage of crossing-over,c) Construct a chromosomal map showing the distance between genes.Parents- Purple,Long PL /pl x Red,Round pl /pl P pp p L l l lGametes (PL) , (Pl), (pL), (pl) (pl)P P p p p L l L l l Offspring-PL/pl ; Pl/pl ; pL/pl ; pl/pl P p P p p p p p L l l l L l l lOffspring- Purple,Long-40, Purple,Round-10, Red,Long-10, Red,Round-40 % of crossing-over = Number of offspring(Recombinant) / Total offspring x 100Number of crossovers = purple,round 10 + red,long 10 = 20 % 0f crossing-over = 20 /100 x100 = 20% . . . . . . . . . . . Answer(b)1% is equal to 1 map unit/ Morgans/ Centimorgan.So, the distance between the two genes is 20 morgans. The linkage map is: P20 L . . . . . . . . . . . Answer(c)15. A female Drosophila fly heterozygous for Gray body(G) and Normal Wings (N) is cros-sed with Black, Vestigial fly. The offsprings are produced in the following number:gray,normal-126; gray,vestigial-24; black, normal-26; and black,vestigial-124;a) Show the parental cross on the chromosomes.b) Indicate whether linkage is present or not.c) Calculate the % of crossing-over.d) Construct a chromosomal map.Parents- gray , normal GN/gn x black, vestigial gn/gnG g g g N n n nGametes- (GN) (Gn) (gN) (gn) (gn) G G g g g N n N n nOffspring GN/gn Gn/gn gN/gn gn/gn Gray,nor gray,ves bla,nor bla,ves 126 24 26 124 = total 300. . . . . . .Answer(a)The example cited is 2PTC(Two point test cross)- In unlinked genes(which followMendel’s 2nd law), the result will be 1:1:1:1. But in this case, the result is not1:1:1:1Hence, the genes are linked. Linkage is present. . . . . . . . . . . . . . . . . . .Answer(b)% of crossover = Number of crossovers / total progeny x 100 = 50/300 x100 = 16.7%. . . . . . . . . . . . . . . . . . . . . .Answer© The distance between the two genes is 16.7 mapunits / morgans.The chromosome map is : G 16.7 NECOLOGICAL INSTRUMENTS1.Hygrometers are instruments used for measuring relative humidity. A simple form of a hygrometer is specifically known as a psychrometer and consists of two thermometers, one of which includes a dry bulb and one of which includes a bulb that is kept wet to measure wet-bulb temperature. Modern electronic devices use temperature of condensation, changes in electrical resistance, and changes in electrical capacitance to measure humidity changes.ApplicationsBesides greenhouses and industrial spaces, hygrometers are also used in some saunas, humidors and museums. In residential settings, hygrometers are used to aid humidity control (too low humidity damages human skin and body, while too high humidity favours growth of mildew and dust mite). The sling or motorized psychrometer is used in meteorology, and in the HVAC industry for proper refrigerant charging of residential and commercial air conditioning systems.Psychrometers are also used in the coating industry. The application of paint is very sensitive to humidity and dew point. With a growing demand on the amount of measurements taken the psychrometer is now replaced by a dewpoint gauge known as a Dewcheck. These devices make measurements a lot faster but are often not allowed in explosive environments.2. Rain gaugesRain gauges are thought to be the most ancient weather instruments, and they're believed to have been used in India more than 2,000 years ago. A rain gauge is really just a cylinder that catches rain. If an inch collects in the cylinder, it means an inch of rain has fallen. It's that simple. Most standard rain gauges have a wide funnel leading into the cylinder and are calibrated so that one-tenth of an inch of rain measures one inch when it collects inside. The funnel is 10 times the cross-sectional area of the tube. Rainfall as low as .01 inches can be measured with this instrument. Anything under .01 inches is considered a trace. This standard rain gauge is shown in the figure.3.Anemometer is a device for measuring the wind speed, and is one instrument used in a weather station. The term is derived from the Greek word anemos, meaning wind. The first known description of an anemometer was given by Leon Battista Alberti in around 1450.Anemometers can be divided into two classes: those that measure the wind's velocity, and those that measure the wind's pressure; but as there is a close connection between the pressure and the velocity, an anemometer designed for one will give information about both.A simple type of anemometer is the cup anemometer, invented (1846) by Dr. John Thomas Romney Robinson, of Armagh Observatory. It consisted of four hemispherical cups each mounted on one end of four horizontal arms, which in turn were mounted at equal angles to each other on a vertical shaft. The air flow past the cups in any horizontal direction turned the cups in a manner that was proportional to the wind speed. Therefore, counting the turns of the cups over a set time period produced the average wind speed for a wide range of speeds.Most widely used for wind-speed measurements is the revolving-cup electric anemometer, in which the revolving cups drive an electric generator. The output of the generator operates an electric meter that is calibrated in wind speed. The useful range of this device is approximately from 5 to 100 knots. A propeller may also be used to drive the electric generator, as in the propeller anemometer. In another type of wind-driven unit, revolving vanes operate a counter, the revolutions being timed by a stopwatch4. Altimeter is an instrument used to measure the altitude of an object above a fixed level. The measurement of altitude is called altimetry, which is related to the term bathymetry, the measurement of depth underwater.A pressure altimeter (also called barometric altimeter) is the altimeter found in most aircraft. In it, an aneroid barometer measures the atmospheric pressure from a static port outside the aircraft. Air pressure decreases with an increase of altitude—approximately 100 hectopascals per 800 meters or one inch of mercury per 1000 feet near sea level.The altimeter is calibrated to show the pressure directly as an altitude above mean sea level, in accordance with a mathematical model defined by the International Standard Atmosphere (ISA). Older aircraft used a simple aneroid barometer where the needle made less than one revolution around the face from zero to full scale. Modern aircraft use a "sensitive altimeter" which has a primary needle that makes multiple revolutions, and one or more secondary needles that show the number of revolutions, similar to a clock face. In other words, each needle points to a different digit of the current altitude measurement.5. Light meter is a device used to measure the amount of light. In photography, a light meter is often used to determine the proper exposure for a photograph. Typically a light meter will include a computer, either digital or analogue, which allows the photographer to determine which shutter speed and f-number should be selected for an optimum exposure, given a certain lighting situation and film speed.Light meters are also used in the fields of cinematography and scenic design, in order to determine the optimum light level for a scene. They are used in the general field of lighting, where they can help to reduce the amount of waste light used in the home, light pollution outdoors, and plant growing to ensure proper light levels.a. Set ISO/ASA of film being used.b. Hold light meter in front of scene with the sphere pointed at the camera.c. Depress center button.d. Needle will move to a reading.e. The reading is measured on the foot-candle scale.Depending on the lighting conditions, there are two settings that can be utilized – the Red Arrow setting (when the High Slide is inserted in the slot below the sphere) is used outdoors in bright light and the Black Arrow setting (High Slide is removed) is used in lower light circumstances.f. Move dial to Black Arrow setting when High Slide is not used so the number lines up with the corresponding number on scale.org. Move dial to Red Arrow setting when High Slide is used so the number lines up with the corresponding number on scale.h. Shutter speed scalei. Aperture scale6.wet-bulb temperature is a type of temperature measurement that reflects the physical properties of a system with a mixture of a gas and a vapor, usually air and water vapor. Wet bulb temperature is the lowest temperature that can be reached by the evaporation of water only. It is the temperature you feel when your skin is wet and is exposed to moving air. Unlike dry bulb temperature, wet bulb temperature is an indication of the amount of moisture in the air. Wet-bulb temperature can have several technical meanings:Thermodynamic wet-bulb temperature: the temperature a volume of air would have if cooled adiabatically to saturation at constant pressure by evaporation of water into it, all latent heat being supplied by the volume of air.The temperature read from a wet bulb thermometerAdiabatic wet-bulb temperature: the temperature a volume of air would have if cooled adiabatically to saturation and then compressed adiabatically to the original pressure in a moist-adiabatic processECOLOGY EXPERIMENTS1.TOTAL ALKALINITY ( CO32- & HCO3-)AIM : To determine the quantity of Carbonates and Bicarbonates in the given water sample.MATERIAL : 250ml conical flask, burette with stand, 0.1 N HCl, Methyl orange indicator, Phenolphthalein indicator.PROCEDURE : 100ml of water sample is taken in a flask and add 4-5 drops of Phenolphthalein indicator to it. If the solution remains colourless, the phenolphthalein alkalinity (PA) is zero, i.e., carbonates are absent and total alkalinity with methyl orange is only determined. If the colour changes to pink after adding phenolphthalein, titrate it with 0.1N HCl until the colour disappears at the end point. This is the phenolphthalein alkalinity. Now , add 2-3 drops of methyl orange to the same sample and continue the titration further till yellow colour changes to pink. This is total alkalinity(TA) ,i.e., the amount of bicarbonates. Calculate the carbonates (PA) and bicarbonates (TA) by using the following formulae.CALCULATIONS : ( AxN) of HCl X 1000 X50 Carbonates(PA) present ( as CaCO3 mg/L) =--------------------------------------- ml of water sample (BxN) of HCl X 1000 X 50 Bicarbonates(TA) present( as CaCO3 mg/L)=------------------------------------- ml of water Where , A= ml of HCl used only with phenolphthalein, B= ml of HCl used with both the indicators (i.e., total HCl used).RESULT: The Carbonates present in the sample water = mg / Lit The Bicarbonates present in the water sample = mg./Lit.2.MECHANICAL ANALYSIS OF SOIL ( Soil texure )AIM : Determination of soil texure by mechanical analysis of the given soil sample.MATERIAL : 100ml measuring cylinder, distilled water, and soil sample.PROCEDURE : Take 20gm of soil in 100ml measuring cylinder. Distilled water is added up to 100 mark. Now shake contents well, and keep the cylinder with out disturbance for 1 hour. Read the volumes of various soil particles and total volume of the soil. The percentages ( by volume ) of various soil particles can be calculated as follows:CALCULATIONS : Total volume of the soil…………..= V1 ( % of soil particle= Volume of that particle x100/ total soil volume) Volume of coarse sand(2.0-20mm).= V2% of coarse sand = V2 X100 / V1 = Volume of fine sand (0.02-2.0mm)… = V3 % of fine sand = V3 X 100 / V1 =Volume of silt ( 0.002-0.02mm)……= V4 % of silt = V4 X 100 / V1 =Volume of clay ( <0.002mm)……… .= V5 % of clay = V5 X 100 / V1 = Measuring cylinder Supernatant soil solution Clay SiltFine sandcoarse sand The super natant clear solution be used for determination of pH. The pH meter electrode may be dipped in the soil solution and directly read the pH. If pH meter is not available, universal indicator ( of BDH) OR pH paper may be used to determine the pH.RESULT : 1st sample- The pH of the soil is………..,hence the soil is Acidic / Basic. 2nd sample- The pH of the soil is………..,hence the soil is Acidic / Basic. ECOLOGICAL SPECIMENS& SLIDES HYDROPHYTES PISTIA: 1. It is a ‘Free Floating Hydrophyte’. 2. These plants float freely on the water surface and are not rooted. 3. There is an offset type of stem running parallel to the surface of water. 4. The leaves arise in rosette manner. 5. The adventitious fibrous roots arise from the nodes of the offsets, and act as balancing roots. 6. The roots have root pockets instead of root cap. 7. The aerenchyma present in inflated petioles , offset stems and roots help in buoyancy EICHORNIA: 1. It is a ‘Free Floating Hydrophyte’. 2. These plants float freely on the water surface and are not rooted. 3. There is an offset type of stem running parallel to the surface of water. 4. The leaves arise in rosette manner, having inflated petioles and a lamina above. 5. The adventitious fibrous roots arise from the nodes of the offsets, and act as balancing roots. 6. The roots have root pockets instead of root cap. 7. The aerenchyma present in inflated petioles , offset stems and roots help in buoyancy. Pistia Eichornia HYDRILLA: 1. It is a ‘ Rooted Submerged Hydrophyte’. 2. The plants are totally submerged under the water and rooted in soil. 3. The stem is slender, herbaceous, green, branched wth nodes and internodes. 4. The adventitious roots arise from the nodes of the prostrate stem. 5. Small, membranous, green leaves are produced from the nodes in whorled manner. 6. Air cavities are present in the stem, as a feature of hydrophyte. Hydrilla Plant Hydrilla stem T.S.Nymphaea stellataIt is also known as water lily or blue lotus is a common flowering water flower, as much popular as lotus in India.1.The aquatic plant has rhizome stem on the bottom of a pond or river and the roots attached firmly2 The leaves and flowers float above water-level with long stalks.3. The leaves are large and rotate. Nymphaea plant Nymphaeapetiole T.S.Vallisneria are members of the monocotyledon family Hydrocharitaceae. This group that should be familiar to aquatic horticulturists as it also contains other important genera; Elodea (anacharis), the beautiful heart-shaped Limnobium and notorious Hydrilla verticillata, among others. The Vallisnerias are superficially similar to the equally popular and appropriate arrowheads, genus Sagittaria. These look-alikes have pointier leaves, whiter, thicker roots and different venation in their leaf structure. Sag's typically bear darker green, stiffer leaves than vals.Vallisneria is a genus of aquatic plant, commonly called eelgrass, tape grass or vallis. The genus has 6-10 species that are widely distributed, but do not grow in colder regions.Vallisneria is a submersed plant that spreads by runners and sometimes forms tall underwater meadows. Leaves arise in clusters from their roots. The leaves have rounded tips, and definite raised veins. Single white female flowers grow to the water surface on very long stalks. Tape grass fruit is a banana-like capsule having many tiny seeds.Sometimes it is confused with the superficially similar Sagittaria when grown submerged.XEORPHYTES ASPARAGUS: 1. It is a ‘succulent xerophytic phylloclade’. 2. It is a small shrub, with green ,much branched, cladode stems. 3. One internodal phyllocldes are known as ‘Cladodes’. 4. The roots are adventitious, tuberous and store water. 5. The leaves are lacking, and the stem is modified into small, green, flattened cladodes, in the form of leaves.AAA Asparagus phylloclades A phylloclade in T.S.Opuntia, also known as nopales or Paddle Cactus from the resemblance to the ball-and-paddle toy, is a genus in the cactus family, Cactaceae.Currently, only prickly pears are included in this genus of about 200 species distributed throughout most of the Americas. Chollas are now separated into the genus Cylindropuntia, which some still consider a subgenus of Opuntia. Austrocylindropuntia, Corynopuntia and Micropuntia are also often included in the present genus, but like Cylindropuntia they seem rather well distinct. Brasiliopuntia and Miqueliopuntia are closer relatives of Opuntia.The most commonly culinary species is the Indian Fig Opuntia (O. ficus-indica). Most culinary uses of the term 'prickly pear' refer to this species. Prickly pears are also known as tuna, nopal or nopales, from the Nahuatl word nōpalli for the pads, or nostle, from the Nahuatl word nōchtli for the fruit; or paddle cactus (from the resemblance to the ball-and-paddle toy). They are native to Mexico. Euphorbia antiquorumIt belongs to the family EUPHORBIACEAESucculent woody shrubs 1 to 3 cm high with milky sap. Mature stem cylindrical in shape with 3 to 6 ridges; younger branches green with 3 to 5 ridges. The projection of the ridges armed with a pair of 2 to 3 mm long spines.· Leaves: few, borne on the ridges, succulent, obovate to oblanceolate to spathulate in shape. Apex obtuse with a small pointed projection, base gradually narrowing downward, sessile. Deciduous.· Flowers: male flower with only 1 stamen, filament short; female flower situated alone at the center of the cyathium, protruding beyond the involucre. Styles 3, not joined to each other, each style forking towards the tip. Yellowish-green, monoecious inflorescence, composed of several staminate (male) flowers and one pistillate (female) flower borne on a green, hemispherical involucre. The whole cuplike cymose inflorescence is called cyathium.· Fruits: glabrous, smooth, about 1 cm in diameter.