Faculty of Science and Mathematics
LABORA LAB ORATOR TORY Y MANUAL MAN UAL
SBU 3023 BIOLOGY II
LABORATORY GUIDELINES AND PROCEDURES SBU3023 BIOLOGY II
ATTENDANCE 1. Attendance at all practical classes is compulsory. Students who fail to attend any of the practical session have to provide a written notification or medical certification. 2. All students who are attending a practical session must sign the attendance sheet, else the student will be considered absent.
PREPARATION 1. Students need to be well prepared and planned their experimental exercise thoroughly before the practical session.
PRACTICAL REPORT Students must write a practical report for each of the practical exercise. All of the report must be hand in by the end of the week of the practical session to the lab. a) Diagram (a practical exercise that involves observing slides through microscopes) b) A practical report must consist of: i) Title ii) Practical exercise iii) Objective iv) Prediction/ Hypothesis v) Results/Findings/Observations (includes any table, graph, description, or diagram) vi) Discussion
Discuss and analyse the results obtained from the practical exercise. Provide a justification of your finding in the practical exercise. vii) Conclusion viii) References All of the references and resources used in the report need to be cited properly.
SBU3023 BIOLOGY II
CELL DIVISION – MITOSIS & MEIOSIS Laboratory exercise 1: Objective:
Investigate the cell divisions in mitosis and meiosis
To understand cell cycle stages via mitosis and meiosis
Introduction:
Organism produces its offspring through the process of reproduction, which involved cells division. Mitosis and meiosis are both cell division mechanisms. However, the outcomes of each mechanism are different. You will study and review the cell division mechanisms at different mitosis and meiosis stages using the Allium cepa root tips and some prepared slides. Material and Apparatus: A. Mitosis
1. Allium cepa root tips ( treated and untreated root tips) 2. Microscope 3. Glass plate 4. Glass slide 5. Pin 6. Cover slip 7. Filter paper 8. Alcohol lamp 9. Acidic Aseto Orsein
Note: The root is treated with paradicholorobenzene in 3 hours
Root tips are preserved by using pure ethyl alcohol and glacial acetic acid (3:1) C.
B. Meiosis
1. Microscope 2. Prepared slides
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SBU3023 BIOLOGY II
Methods:
1. Place 3 or 4 root tips in several drops of acidic Aseto-orsein (9 acidic Aseto Orsein: 1 10% HCl) on the glass plate. 2. Heat the glass plate by using alcohol lamp for 2-3 minutes. (Careful not let the glass plate to burn over). 3. Cover with the glass plate and leave it for 10 minutes. 4. Place a root tip on a clean glass slide. Cut into 1 to 2 mm from the end of the tip and remove the remaining part. 5. Squash the root tip to the tiny pieces by using a pin. 6. Add 1 or 2 drops of Aseto-orsein onto the slide. Carefully wipe any remaining Aseto-orsein around the specimen with filter pape r. 7. Cover the glass slide with a cover slip (do not allow it to dry), then gently knock the cover slip by using a short piece of wood (matches). This step is performed to separate the root cells. 8. Slowly heat the glass slide by placing it on the alcohol lamp for a while. (Careful not let the glass slide to burn over). Then, place the glass slide between filter papers and press the cover slip gently for a complete cells separation. 9. Examine the slides under the microscope.
Note: Apply step 4 to step 9 on treated and untreated root tips. Apply step 9 on the prepared slides of the meiosis stages.
Observation:
Observe the slides and identify the cell division stages that you can see. Describe and draw each of the observed stages in your report.
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SBU3023 BIOLOGY II
MENDEL’S INHERITANCE LAW Laboratory exercise 2:
Principle of Mendel
To understand the concept of the Mendelian’s Inheritance Laws
Objective: Introduction:
Gregor Mendel was the first person who studied the genetic materials transmission in inheritance. He found the inheritance principles that were the basic of modern genetics. These inheritance principles were also known as Mendel’s First Law: Segregation, the Second Law: Independent assortment. Mendel’s Law of Segregation: During the gametes formation, the two alleles of each trait separate (segregate), and then unite at random, one from each parent, at fertilisation. Mendel’s Law of Independent assortment: During gametes formation, different pairs of alleles segregate independently from each other.
Material and apparatus:
1. Maize kernels a. Ratio 3:1 b. Ratio 1:1 c. Ratio 9.3.3.1 d. Ratio 1:1:1:1 Methods: A. Principle of segregation
1. You will be given maize kernel, which resulted in a cross of the contrasting traits. Observe the characteristics appeared on the maize kernels (a) and (b). Observe the characteristics appeared on the maize kernels (a) and (b). Maize kernel (a) is obtained from F1 cross which was cross between purple seed and yellow seed.
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SBU3023 BIOLOGY II
While, maize kernel (b) is obtained from a test cross of F1 generation (purple X yellow). 2. Determine the dominant phenotype 3. Count the number of every phenotypes 4.
Count the expected number of every phenotype
B. Principle of independent assortment
1. You will be given the maize kernels, which resulted from a cross of two pairs of contrasting traits. Observe the characteristics appeared on the maize kernels (c) and (d). Maize in kernel (c) is obtained from the F1 cross which was a cross between purple and round seed with yellow and wrinkled seed. While maize kernel (d) is obtained from a test cross of F1 generation, (purple and round seed X yellow and wrinkled seed). 2. Determine the dominant and recessive phenotype 3. Count the numbers of every phenotype 4. Count the expected numbers of every phenotype Observation:
Record your observation in a table form. Review and write a report about your observation and understanding on the Mendelian’s Laws.
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SBU3023 BIOLOGY II
From DNA to Protein Laboratory exercise 3:
DNA, mRNA and protein.
To build and appreciate the DNA structure.
Objectives:
To apply and determine the product of transcription and translation processes.
Introduction:
Deoxyribonucleic acid (DNA) holds heredity information. A DNA molecule consists of two long chains of nucleotides that coiled into a double helix. The chains composed of nucleotides, which each has phosphate groups, a deoxyribose sugar and a nitrogenous base. These two chains held together by the hydrogen bonds. The DNA puzzle kit includes the following and should be carried out in the following order: 1. DNA: the genetic code 2. The Code Transcribed and Translated
A. DNA: The genetic code
Materials and apparatus:
24 deoxyribose units (red) 24 phosphate units 4 adenine units 8 cytosine units 8 guanine units 4 thymine units
Experimental procedures:
I. Constructing DNA nucleotide 1. Attach one phosphate unit to the free bond from CH2 deoxyribose molecule. 2. At the right hand side of the oxygen atom, there is a carbon atom with one free bond. Attach the adenine to the bond. The structure you have constructed is called adenine nucleotide. 1
SBU3023 BIOLOGY II
3. Construct nucleotide having guanine, cytosine and thymine.
II. Joining nucleotide through covalent building 1. Select an adenine nucleotide and a guanine nucleotide that had been prepared earlier. 2. Attach both nucleotides together at phosphate unit from one nucleotide to 3’ end of another nucleotide. 3. Repeat step 2 until you get a DNA chain consisting of six nucleotides.
III. Base pairing of DNA molecules 1. Separate the DNA chains constructed approximately 30cm apart. 2. Built two new DNA chains on each separated stand, which complement and anti-parallel with each other.
B. The code transcribed and translated
Materials:
12 deoxyribose units (red) 12 ribose units (pink) 24 phosphate units 4 adenine units 8 cytosine units 8 guanine units 2 thymine units 2 uracil units
Experimental procedures:
1. Construct a DNA chain having the following sequences: CGT CCA CGT CCA 2. Construct a RNA chain with is complement and antiparallel with the DNA chain built in step 1. Remember, in RNA base thymine is replaced by base uracil. 3. By referring to ‘codon table for mRNA’, fill in the table below with their respective translated mRNA sequences and transcribed amino acids. 2
SBU3023 BIOLOGY II
DNA sequence
Codon mRNA
in
Amino Acid
CGT CCA CGT CCA
1. What are the differences between DNA and RNA? 2. Which direction does DNA replication occurs? 3. What are the differences between a leading and lagging strand in DNA replication? 4, State the base sequences of start codon and stop codon.
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SBU3023 BIOLOGY II
TAXONOMY: ORGANISMS CLASSIFICATION Laboratory exercise 4: Objectives:
Classify invertebrate into the taxa
to determine organisms using the dichotomous key.
Introduction:
Dichotomous key is one of a common method used to classify an unknown organism. The observation on the characteristics, such as structure, behaviour, of the unknown organism helps to recognise the organism. However, careful observations are required to induce the right name of the organism. “Dichotomous” means “divided into two parts”. Therefore, dichotomous keys always offer two choices for each steps. Each key describes a characteristic of a particular organism or group of organisms.
Materials
1. Prepared slides 2. Microscope
Procedures:
1. Examine the prepare slides using the microscope. Examine your material first using the lower power objective (i.e. 10X); then use a higher power objective (i.e. 20X or 40X). Because the objectives are parfocal, you need to use only the fine focus knob to fine tune your image. Never use the coarse adjustment t o focus downward. Replace and remove a slide only after the lowest power objective has been rotated into viewing position. 2. Use the dichotomous key shown in the Appendix 1 (Darley M., 2003) to recognise the organisms. 3. Record the name of the organisms and draw the features of each organism from your observation. Note the power objective of the microscope used in the observations.
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SBU3023 BIOLOGY II
Appendix 1 A Dichotomous Key of Pond Life
1a 1b
Organism is a filament (a liner series of cells) – 2 Organism is a unicell or a colony (flat or spherical) – 6
2a 2b
Filament is unbranched – 3 Filament is branched – Chaetophora (Cholophyceae)
3a 3b
Choloplast fills cell: filaments often end in “H” piece – Microspora (Cholophyceae) Chloroplast has a distinctive shape – 4
4a 4b
Chloroplast is in the form of a spiral – Spirogyra (Charophyceae) Chloroplast is not in the form of a spiral – 5
5a 5b
Two “star -shaped” chloroplast in each cell – Zygnema (Charophyceae) Two thin, “plate-like” chloroplasts in each cell – M ougeotia (Charophyceae)
6a 6b
Organism is a unicell – 7 Organism is a colony – 13
7a 7b
Organism cell does not contain flagella – 8 Organism cell is motile with flagella – 11
8a 8b
Organism cell is green – 9 Organism cell is golden brown and elongated (pinnate diatom) – Nitzschia (Strminopila)
9a 9b
Cell is round, not divided into halves – Chlorococcum (Chlorophyceae) Cell is not round, often appears to be divided into halves with nucleus in the middle of the cell, dark green (desmids, Charophyceae) – 10
10a
Cell is elongated without a constriction in the middle; cell may be slightly curved (like a banana) – Closterium Cell has an obvious constriction in the middle; cell highly ornate with several lobes and secondary lobes – M icr asteri as
10b
11a 11b
Cell is green – 12 Cell is not green, with two sub-apically inserted flagella; may be blue-green, brown, radish brown – cryptomonads (genus unknown)
12a 12b
Cell elongated with one long flagellum – Euglena (euglenid, Euglenozoa) Cell is oval with two flagella – Chlamydomonas (Chlorophyceae) 2
SBU3023 BIOLOGY II
13a 13b
Colony is spherical and motile – 14 Colony is non-motile – 15
14a 14b
Colony is green – Volvox (Chlorophyceae) Colony is yello-brown – Synura (Chrysophyta)
15a 15b
Colony is spherical – Coelastrum (Chlorophyceae) Colony is not spherical – 16
16a 16b
Colony is a round, flat plate – Pediastrum (Chlorophyceae) Colony has 2 or 4 cells with spines on the corners – Scenedesmus (Chlorophyceae)
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