School Science Lessons
Biology experiments
2009-10-19
Please send comments to: J.Elfick@uq.edu.au
Table of contents
4.4.0 DNA and RNA
9.4.0 Mitosis and meiosis
9.6.0 Drosophila experiments, Mendel's laws
9.254.0 Human genetics
4.4.0 DNA and RNA
4.4.1 Isolate DNA from any living material
4.4.2 Isolate DNA from wheat germ, kiwi fruit, strawberries
4.4.3 Isolate DNA from cheek cells
4.4.4 Isolate DNA from sweetbread (calf thymus gland)
9.4.0 Mitosis and meiosis
9.4.1 Mitosis in cells of onion root tip
1.13.0A Ephedrine group
9.57 Cells and tissue sections, T.S., L.S., R.L.S., T.L.S
9.9.8 Meiosis in grasshopper testes
9.6.0 Drosophila experiments, Mendel's laws
9.6.1 Cultures of fruit fly (Drosophila melanogaster) for heredity experiments
9.6.2 Fruit fly strains
9.6.3 Mendel's experiments
9.6.4.1 Mendel's first law, law of uniformity (inheritance of one pair of characteristics)
9.6.4.2 Mendel's first law, law of uniformity (inheritance of two pairs of characteristics)
9.6.5 Mendel's second law, law of segregation
9.6.6 Mendel's third law, law of independent assortment
9.6.7 Sex-linked inheritance, introduction of recessive characteristic by female
9.6.8 Sex-linked inheritance, introduction of recessive characteristic by male
9.6.9 Genetics, lethal factors
9.6.10 Giant chromosomes
9.254.0 Human genetics
9.254 Human genetic inheritance
9.254.1 Human traits with simple inheritance patterns
9.254.2 PTC tasters and non-tasters
9.254.3 Unattached / attached earlobe
9.254.4 Fingerprint variations
9.4.1 Mitosis in cells of onion root tip
See diagram 9.107.1: Mitosis in onion root tip cells | See diagram 9.107.2: Cell division in onion root tip | See diagram 2.26: Microscope technique, staining
The fundamental reproductive process called cell division may be studied by selecting an appropriate tissue that is growing rapidly. A good source of such cells is the root tip region of onions or other related plants. Onion bulbs, garlic cloves or onion sets placed in an aerated water bath provide large quantities of material.
The cells of growing and developing organisms are constantly multiplying by cell division. In unicellular organisms daughter cells form which separate after the division process is complete and continue to develop independently. In multicellular organisms the cell mass is increased by cell division. The organisms grow. In cells without a nucleus cell division takes place using a simple constriction process. However, the division process in nucleated cells involves a complicated mechanism of division of the cell nucleus, during which operation the chromosomes are clearly visible and are distributed between the two daughter cells after longitudinal splitting. This process is called indirect nuclear division or mitosis. Observe the different phases of indirect nuclear division, mitosis.
In the onion root tip, before mitosis starts, interphase, granules in the nucleus have the staining reaction of chromatin that consists of DNA and proteins, mostly histones. During interphase DNA replication occurs so the chromosome appearing in prophase will have two identical sister chromatids.
Prophase: The loose coils of the chromatin network condenses to become the 16 chromosomes of the onion cell, double structures with two identical chromatids joined by a centromere. The nuclear membrane and nucleoli disappear. Prophase occupies about two thirds of the time taken for mitosis.
Metaphase: Curved microtubules, fibres, form a spindle outside the nucleus. The centromeres move to the central plane of the spindle, metaphase plate. The centromeres split in two.
Anaphase: The identical chromatids move to the opposite poles of the spindle to leave two identical sets of chromosomes and a new nuclear membrane appears around each set. The chromatids are now chromosomes.
Telophase: Nucleoli reappear. The chromosomes return to being chromatin granules.
Cytokinesis is the division of the cytoplasm between the new nuclei and the formation of new cell walls.
Mitotic cell division is the basis of growth and tissue repair in multicellular organisms. Mitosis keeps the chromosome number constant.
1. Cut off the white root tips of healthy specimens. Cut a 3 mm cylinder from the end of a root. Put it in a drop of aceto-carmine stain on a microscope slide. Cut up the onion tip with a razor blade until the pieces are extremely small. Cover the preparation with a cover glass. With a piece of folded paper towelling over your thumb for protection, gently squash the pieces of root tip by pressing on the coverslip with a rolling motion. Do not allow the cover glass to slide. Then examine the preparation with the low power of a microscope. Look for dark stained threadlike bodies. These are chromosomes or mitotic figures. Find the find various types or stages and count the number of various stages. From this information estimate the relative lengths of time that the various stages are present in a reproducing cell.
2. Plant an onion or shallot in moist absorbent paper in a warm place to obtain roots. Cut off 1 cm lengths from the ends of roots and fix them in a solution of 1 part glacial acetic acid to 3 parts 95% alcohol. Leave for 24 hours. Put a piece of root in a drop of aceto-carmine on a slide. Cut off 3 mm of the tip and discard the rest. Gently warm over a spirit lamp. Place a coverslip over the drop of stain and apply gentle pressure to separate the cells. These cells will show stages in mitosis.
3. Put onion root tips in 1 mL orcein in a watch glass. Heat over a spirit lamp for 1 minute or until the tips are soft. Scrape the tips over a microscope slide to make an even smear. Add drops of glycerine and a coverslip.
4. Put an onion in moist absorbent paper in a warm place to obtain roots. Cut off 1 cm lengths from the ends of roots and fix them in a solution of 1 part glacial acetic acid to 3 parts 95% alcohol. Leave for 24 hours. Put a piece of root in a drop of aceto-carmine on a slide. Cut off 3 mm of the tip and discard the rest. Gently warm over a spirit lamp. Place a coverslip over the drop of stain. Wrap absorbent paper over the thumb, then gently squash the pieces of root tip by pressing on the coverslip with a rolling motion. Be careful! Do not break the coverslip because the broken pieces are very sharp and dangerous. Do not allow the coverslip to slide. These cells will show stages in mitosis.

5. In the morning, cut 5 mm from the end of a growing root of onion or pea. Cut the piece of root twice longitudinally. Put the longitudinal sections in a drop of carmine acetic acid on a microscope slide. Cover with a coverslip and heat to boiling point holding the slide with forceps and holding it over a small Bunsen burner flame by moving the slide backwards and forwards to prevent excessive heating in one place. Put a drop of 2% acetic acid at the side of the coverslip and draw it across under the coverslip with absorbent paper on the opposite side. Press down on the coverslip with a scalpel handle to squash the cells. Be careful not to break the coverslip as the broken pieces are very sharp. Examine the cells under high power. Look for dividing nuclei at different stages of development, prophase, metaphase, anaphase and telophase. Count the number of chromosomes, e.g. onion 16 and pea 14. Find the different stages and count the number of separate stages. Use this information to estimate the relative lengths of time for each stage during mitosis. The process takes about two hours with prophase taking two-thirds of the time. Repeat the experiment with root tip specimens taken at different times.

6. Plant an onion or shallot in moist absorbent paper in a warm place to obtain roots. Cut off 1 cm lengths from the ends of roots and fix them in a solution of 1 part glacial acetic acid to 3 parts 95% alcohol. Leave for 24 hours. Put a piece of root in a drop of aceto-carmine on a slide. Cut off 3 mm of the tip and discard the rest. Gently warm over a lamp (light) bulb. Place a coverslip over the drop of stain and apply gentle pressure to separate the cells. These cells will show stages in mitosis.

7. Examine a prepared slide showing cells in various stages of mitotic division, e.g. a stained longitudinal section through the root tip of the onion.
9.9.8 Meiosis in grasshopper testes
See diagram.9.108.1: Meiosis 1 | See diagram 9.108.2: Meiosis 2 | See diagram 9.108.3: Meiosis models A | See diagram 9.108.4: Meiosis models B | See diagram 9.108.5 Meiosis models C
Meiosis can be seen in grasshopper testes or in Ascaris lumbricoides, where the eggs remain dormant until fertilized. During fertilization the male and female gametes fuse to form a zygote so the number of chromosomes doubles, n chromosomes from the male gamete and n chromosomes from the female gamete. The 2n cells in the sex organs undergo meiosis to halve the number of chromosomes to n chromosomes in the male gamete or female gamete.
Prophase I Leptotene: The chromosomes appear like a long string with of beads, called chromomeres.
Prophase I Zygotene: The homologous chromosomes from the male and female gametes lie closely alongside each other forming shorter and thicker paired structures so appearing to halve the number of chromosomes. They are attached to the inside wall of the nucleus in animals but form a tangled heap in plants.
Prophase I Pachytene: When the pairing complete and the chromosomes appear as "thick" strings. The two chromosomes are referred to as a "bivalent," while the same structure viewed as four chromatids is known as a "tetrad."
Prophase I Diplotene: Each chromosome of a pair splits into two chromatids along their length except at the centromeres. As the chromosomes separate they remain attached at chiasmata where recombination of the genes occurs.
Prophase I Diakinesis: The homologous chromosomes repel each other and continue contracting. The nucleolus and nuclear membrane disappear.
Metaphase I: The bipolar spindle appears and the paired chromosomes (tetrads) line up on the metaphase plate and attach to the spindle fibres with the centromeres.
Anaphase I: Each homologue moves toward opposite poles so halving the number of chromosomes. This process is the reduction division that characterizes meiosis.
Telophase I: The chromosomes may become surrounded by a nuclear membrane but usually the second division starts when the chromatids of each chromosome separate and move to the opposite pole of the spindle.
Second division: This division is similar to normal mitosis except that the number of chromosomes is n not 2n. The four n nuclei develop nuclear membranes during the telophase of the second division to produce a tetrad of pollen grains but only one n nucleus develops in the ovule.
9.254 Human genetic inheritance
See diagram 9.254: Darwin's point, unattached earlobe, widow's peak
Members of a family are always interested in "family resemblance", how members of the same family look alike. The resemblance is due to traits inherited from parents to offspring. The "expression of traits" (phenotype) depends on an individual's "genetic make-up" (genotype). Human genes are carried on 23 pairs of chromosomes. One set of 23 chromosomes is inherited from the mother and the other set of 23 chromosomes is inherited from the father. About 10 000 genes are carried on 44 of the 46 chromosomes. The other two chromosomes are the X and the Y chromosome. Females have two X chromosomes and males have an X and a Y chromosome. The X chromosome is similar to the other chromosomes but the Y chromosome is much smaller with only about 30 working genes that control maleness. Each trait may be controlled by a single pair of genes, i.e. one of each gene pair (allele) inherited from each parent, or controlled by multiple pairs of alleles, i.e. many gene pairs inherited from each parent. Some traits are high frequency traits. Some traits depend on the sex of a person. For example, index finger (second finger) shorter than ring finger (fourth finger) is the dominant phenotype only in males but index finger longer than ring finger is the dominant phenotype in females. Human DNA contains about 25 000 genes. However, we also have some genes in the cell mitochondria. Mitochondrial DNA (mtDNA, mDNA) control the Kreb's cycle in our basic metabolism. Sperm have a few mitochondria in the middle piece but the ovum has about a quarter of a million mitochondria. After fertilization, the mitochondria from the sperm all die so we inherit only mitochondria genes from the ovum of the mother giving us the same mDNA as our mother had. Mitochondrial DNA forms in a circle of 6,589 DNA bases. Mitochondrial DNA mutates much faster than nuclear DNA so that almost everyone has different mitochondrial DNA but relatively similar nuclear DNA. Some mutations of mitochondrial DNA can cause serious problems, e.g. exercise intolerance, Alzheimer's disease, diabetes.
9.254.1 Human traits with simple inheritance patterns
See diagram 9.254.1: Fingers and toes
Some traits may not be due to a single pair of dominant or recessive genes. For example, brown eye genes are dominant over green eye genes and both are dominant over blue eye genes. However, eye colour is caused by many genes that control the colour and the amount of melanin pigments in the iris. The child of both parents with light colour eyes may have dark colour eyes due to mutations and other changes in the chromosomes. Eyes may be darker after exposure to light and mutations can cause unusual eye colours. Similarly, hair colour is caused by multiple genes and may change with age.
.
 Dominant Recessive
1.1
 Unattached earlobe, free earlobes Attached earlobe (to the face)
1.2
 Big toe shorter than second toe Big toe longer than second toe
1.3
 Widow's peak (downward forehead
V-shaped peak of hair)
No widow's peak
1.4
 Darwin's point on the ear, skin flap
No Darwin's point skin flap
1.5
 Hair on second joint of any finger
on either hand (middle segment)
No hair on any middle segments of fingers
1.6
 Straight thumb, cannot bend thumb back 50o "Hitchhiker's thumb"
1.7
 Little finger bends towards 4th finger, crooked pinkie
 Little finger straight, pinkie straight
1.8
 Facial dimples, cheek dimples No facial dimples when you smile
1.9
 Freckles
No freckles
1.10
 Can roll tongue to form U-shape Cannot roll tongue
1.11
 Hair whorl clockwise Hair whorl anticlockwise
1.12
 Can taste PTC Cannot taste PTC, non-taster
1.13
 Left thumb on top of interlocking
fingers
Right thumb on top of interlocking fingers
9.254.2 PTC tasters and non-tasters
See 1.13 Phenylthiocarbamide, PTC, phenylthiourea, PTU, C7H8N2S
Be careful! Phenylthiourea (PTC, PTU, phenyl thiocarbamide, n-phenylthiourea, structural formula: C6H6NHCSNH2) is very toxic and may be fatal if swallowed.
1. The ability to taste phenylthiourea (bitter) and other compounds with an –N-C= group is an autosomal dominant trait, i.e. the defective gene is dominant and will be inherited by half the offspring of either sex. Phenylthiourea tasters detect saccharin, potassium chloride (KCl) and caffeine as more bitter.
2. Put a piece of water-soaked absorbent paper on the tongue and roll it around the tongue for 10 seconds. Repeat with absorbent paper soaked in PTC (phenylthiourea). Be careful! Do not swallow the PTC. Rinse the mouth out with water. Did you notice a bitter taste?
9.254.3 Unattached / attached earlobe
Investigate the sameness and difference between your parents and you for the above genetic characteristics. Perhaps all your siblings and one of your parents have the same unusual genetic traits because this characteristic "runs in the family". Some people have an unattached earlobe on one ear and attached earlobe on the other ear.
9.254.4 Fingerprint variations
Look at the tiny ridges on the pad of your fingers. When these are inked and placed on paper, a fingerprint is made. Put the stamp pad near the edge of a table and, in turn, roll your fingers over the pad. Each time, make a fingerprint on your notebook. Label the fingerprint with the number of the finger and whether it is the right or left hand.
4.4.0 DNA and RNA
See diagram 16.3.2.8.1: DNA molecule | See diagram 16.3.2.8.2: Ribose, deoxyribose, nucleotide
See diagram 16.21.10: Purines to adenine, guanine | See diagram 16.21.13: Pyrimidines to cytosine, thymine, uracil
1. DNA is the master molecule that carries all of the inherited characteristics (genes) of an individual in the form of chromosomes. Each individual (such as a human) receives one haploid set of 23 chromosomes from their father's sperm and one haploid set of 23 chromosomes from their mother's egg. The two sets come together at conception when the diploid zygote (fertilized egg) is formed. Each eukaryotic chromosome (the chromosomes of algae, fungi, plants and animals) carries thousands of genes, about 100 000 functional genes per cell. A chromosome is analogous to a high capacity storage disk (DVD disk) while genes are analogous to files on this storage disk. If a chromosome could be completely unravelled, it would reveal a long, ladder-shaped DNA molecule that is coiled into helical spirals. At intervals along this double helix, the DNA ladder is wrapped around small beads of protein called nucleosomes. There are also extra chromosomal genes in the form of bacteria-like (prokaryotic) plasmids within cytoplasmic organelles called mitochondria and chloroplasts. The uprights of the DNA ladder are alternating 5-carbon sugars (deoxyribose) and phosphates. The rungs of the ladder are nitrogenous base pairs, purine bases adenine and guanine and pyrimidine bases cytosine and thymine with adenine always pairing with thymine and guanine always pairing with cytosine. DNA and RNA have nucleotide sub units consisting of a phosphate, a sugar and a base. Every base pair has four different arrangements: A-T, T-A, C-G and G-C that allows millions of different possible arrangements in a DNA molecule. Tiny amounts of DNA can be cloned into millions of copies with the PCR technique (Polymerase Chain Reaction) to give enough DNA to sequence gels into banding patterns to represent different base pair sequences and determine genetic "fingerprints" for crime identification or show relationships among plant and animal species. The results can be compared with gene sequences in gene bank databases.
2. The structure of DNA was first suggested by J. D. Watson and F. H. C. Crick in a paper published in Nature, April 25, 1953, "MOLECULAR STRUCTURE OF NUCLEIC ACIDS, A Structure for Deoxyribose Nucleic Acid", based on the X-ray evidence from research done by Rosalind Franklin and M. Wilkins.
The basic unit of DNA, the nucleotide, contains a sugar molecule (the aldopentose deoxyribose) a phosphate molecule and an organic nitrogenous base, a purine [adenine (A) or guanine (G)] or a pyrimidine [cytosine (C) or thymine (T)]. So DNA consists of a long string of nucleotides. The structure of DNA has two helical chains with each chain coiled around the same axis with the sequence of the atoms in the two chains running in opposite directions. In the diagram, the two ribbons represent the two phosphate to sugar chains and the horizontal rods represent the pairs of bases (base pairs) connecting the chains. Only specific pairs of bases can bond together to adenine with thymine, and guanine with cytosine. The protein-coding sequence of a single human gene may contain more than 1000 base pairs. There are more than 30 000 different human genes. RNA is similar to a single strand of DNA except that uracil (U) substitutes for thymine (T). A codon is the sequence of three consecutive nucleotides in messenger ribonucleic acid (mRNA) which controls the placing of a particular amino acid into a protein or acts as the start or termination signal of a particular protein synthesis. As there are 64 possible codons, i.e. 4 X 4 X 4 and 20 amino acids, so more than one codon can code for a particular amino acid, e.g. the code for arginine is CGC and CGA and CGG. Just before cell division the DNA double helix unwinds to allow each strand to be copied to form a sort of mirror image, pairing A with T and G with C. The result is two DNA molecules each containing one old strand and one new strand with base pair sequences the same as the original DNA.
4.4.1 Isolate DNA from any living material
You can use strawberries, broccoli, split peas, spinach leaf, lettuce, onions or chicken liver.
Put in a kitchen blender one half cup (100 mL) of extraction material, one eighth teaspoon (less than 1 mL) of table salt and 1 cup (about 200 mL) of cold water. Run the kitchen blender on high for 30 seconds or until there is a runny pea soup consistency. Pour this blended material through a strainer into another container. Add 2 tablespoons (about 30 ml) of liquid detergent to the strained mixture and swirl to mix. Let the mixture stand for 5-10 minutes. Transfer the mixture into test-tubes one third full. Tilt the test-tube and slowly pour cold ethanol into the test-tube down the side so it forms a layer on top of the blended mixture. Observe the DNA rising into the alcohol layer from the blended mixture. Use an opened "slide-on" paper clip to draw the DNA into the alcohol..
4.4.2 Isolate DNA from wheat germ, kiwi fruit, strawberries
1. Put wheat germ, or kiwi fruit liquidized in a food processor, in the test-tube to a depth of 1 cm. Add 3 mL of warm water and a drop of kitchen detergent. Plug the test-tube with cotton wool and roll it in the hands gently for three minutes. Float 2 mL of cold methylated spirit on top of the mixture and leave to stand. The cloudiness at the interface of the liquids is the DNA. Use an opened "slide-on" paper-clip to lift out the threads of DNA.
2. Use two large frozen strawberries in a plastic bag and cooled rubbing alcohol [(CH3)2CHOH, isopropyl alcohol, 2-propanol]. To prepare the extraction solution, slowly add 50 mL of kitchen detergent to 250 mL of water so that no froth forms. Slowly add 10 g of table salt (sodium chloride), then mix the solution. Add water to make up the solution to 500 mL. Add 20 mL of the extraction solution to the strawberries and seal the plastic bag after expelling the air. Press down on the plastic bag for 5 minutes to crush the strawberries to a pulp. Put the plastic bag on crushed ice and press down on the plastic bag again.. Filter the mixture of pulped strawberries and extraction solution into a beaker. Put 2 mL of the filtrate into a test-tube and slowly add the same volume of cold rubbing alcohol. Use a glass rod to remove the fluffy strands of DNA from the alcohol layer of the mixture.
4.4.3 Isolate DNA from cheek cells
Dissolve half a teaspoon of salt in half a cup of water. Add a little dishwashing liquid to break up the cells and release the DNA.
Put 25 mL of water in your mouth but do not swallow it. Move the water around your cheeks vigorously for 30 seconds to remove some cheek cells. Spit the water into a clean cup. 2. Add 5 mL of this fluid to a 20 mL test-tube. Add 2.5 mL of the salt and dishwashing liquid solution. Put a stopper on the test-tube and move the test-tube up and down 3 times gently so that the contents do not form froth. This movement breaks up the hundreds of cheek cells to release the DNA from the nucleus. Gently pour into the test-tube a teaspoonful of ice cold ethanol that has been in a freezer for hours before the experiment. Watch the point where the two layers meet. Note the strands of DNA forming as cloudy filaments stretching up into the top ethanol layer. DNA is not soluble in ethanol, so when the ethanol meets the DNA solution it starts to precipitate form a DNA salt. Use a glass hook or [plastic tie-wire to remove strands of DNA by slowly dipping up and down through the two layers.If you then gently invert the test-tube several times until the alcohol is mixed in the precipitated DNA will look like a small ball of white thread.
4.4.4 Isolate DNA from sweetbread (calf thymus gland)
Cells of the calf thymus gland possess large nuclei from which large amounts of DNA can be obtained. Buy some freshly-killed sweetbread from a butcher. Deep-frozen samples also produce comparable results. Each tensides molecule of the kitchen detergent has a hydrophilic and a lipophilic end. When they are taken up, the field of charge of molecules of the lipoprotein membrane is shifted, and hydrolysis occurs. The membranes of the cell and nucleus are destroyed and histone molecules split off from DNA strands. After the addition of kitchen detergent, the suspension becomes clear and viscous. The viscosity is caused by liberated DNA molecules.
1. To produce a thymus suspension, cut up five sweetbreads in a mortar using a pair of scissors. Add clean washed sand to the tissue in the mortar and grind with the pestle. Add 20 mL of tap water and continue grinding until the water is completely clouded. The water is clouded by cells and abraded cell detritus suspended in the water. Filter the suspension through gauze or two layers of surgical gauze.
2. To liberate DNA, add 5 to 10 drops of kitchen detergent to the cell suspension and shake thoroughly until the suspension becomes clear and viscous.
3. To obtain DNA, cover the suspension with one and a half times the amount of cold methylated spirit solution or ethanol solution in a narrow glass beaker. Use a glass rod to remove the DNA precipitating as whitish strands. Rotate the rod slowly and carefully dip it up and down occasionally to mix the suspension and the alcohol. This action will precipitate more nucleic acid. If the strands slip from the rod, use forceps to retrieve them.
9.6.1 Cultures of fruit fly (Drosophila melanogaster) for heredity experiments
See diagram 15.5.1: Drosophila experiments
1. Attract fruit flies by putting overripe fruit in an open container, e.g. a glass jar. After trapping the fruit flies, transfer them to small containers containing fruit chunks, e.g. banana. Put a slice of ripe fruit in the bottom of the container and make a paper funnel with a hole in the end to fit the mouth of the container. Put the container in the open air. When six or eight fruit flies have entered (including both males and females) remove the funnel and plug it loosely with cotton wool. The females are larger, with a broader abdomen. The males are smaller and have a black-tipped abdomen. Soon eggs will be deposited, and in 2 or 3 days the larvae will hatch. Put a piece of paper in the container for the larvae to crawl on when they are ready to pupate. The adult insects will come from the pupae. Put newly-hatched fruit flies in another container to start a new generation. To study fruit fly cultures, cut a piece of graph paper and stand it upright in the container so that you can sample a large population in the bottle by counting the number of pupae on the grid. Make daily counts of the eggs, larvae, pupae and young adults of the population in a bottle. Draw a graph to show the increase in population with time. Maintain the culture for as long as the flies continue to survive.
2. Use conical flasks for breeding. Prepare the artificial diet the day before. Add 2 tablespoons beet juice syrup to 1 litre water. Heat the mixture and stir constantly. Add semolina until a thick paste forms. Prevent moulds forming by stirring in 1 spatula tip of nipagine per litre. Pour the mixture into clean breeding flasks to a depth of 2 cm. After cooling, add 5 drops of a viscous suspension of bakers' yeast in tap water. Close the breeding flasks with cotton wool plugs. Before the flies are inserted, absorb any liquid collected on the surface of the feeding mixture with strips of filter paper to prevent the flies sticking to the surface. Transfer the flies by tapping the breeding flask on the palm of the hand so that the flies fall to the bottom. Quickly remove the cotton wool plug and place a collecting tube with the same size neck as the breeding glass on top of it. By lightly tapping and shaking the flask, you can get the flies to enter the collecting tube. Separate the two flasks and seal with cotton wool plugs. Apply diethyl ether to the plug of the collecting tube so that the flies are anaesthetized within 20 seconds. Shake the flies out on a sheet of filter paper and separate the sexes. Use 8 females and 15 males for each fresh breeding batch. The males are smaller than the females and the shapes of the abdomens differ. The abdomen of the female is larger, more pointed and has 4 or 5 black transverse rings. The abdomen of the male is smaller, more rounded with a black tip and has only two transverse rings. The male has a row of bristles on the first foot section of each front leg. To prevent the fruit flies sticking to the feeding mixture, transfer them to the breeding flask in small cones made from filter paper. Make the cone by twisting a piece of filter paper about 5 cm square around the end of a pencil. If unfertilized females are required, remove all the fruit flies from a breeding container containing a lot of pupae on the point of hatching and after 6 hours collect the fruit flies which have hatched. Since the males are unable to copulate until 8 hours after hatching, the females among them cannot be fertilized. Larvae obtained in the following manner are most suitable for preparing giant chromosomes. Breeding colonies should not be overpopulated so remove adults from the breeding glass after they have deposited their eggs. When the larvae are half grown, add more drops of viscous yeast suspension to the container and remove the cultures to a cool place, 15o9. Take the fully grown larvae which have crept up the glass wall just before pupation for use as specimens.
3. Drosophila medium is an artificial diet consisting of 20 g of agar, 135 g of sugar, 38 g of yeast, 0.12 g of nipagine (10 g L- 1 nipagine in 70% ethanol) made up in 1 L of water and incubated at 20°C.
9.6.2 Fruit fly strains
Different varieties are called mutants because they have developed by mutation of genes. Describe the following strains: Normal wild, vestigial vg, ebony e, white w, and Curly Cy. Anaesthetize the fruit flies with diethyl ether. Put them on a sheet of filter paper and examine them with a 6 X magnifying glass or stereoscopic magnifier. Differentiate between the strains by means of their physical characteristics, e.g. colour of body and eyes. Compile a table of observations:
Strain characteristic Normal,
wild type
+		 vestigial
vg		 ebony
e		 white
w		 Curly
Cy
Colour of body .
 . . . .
Colour of eyes . . . . .
9.6.3 Mendel's experiments
Gregor Mendel (1822-1884) used garden peas (Pisum sativum) for most of his experiments because they have constant different characters, flowers of hybrids can be protected from all other pollen and hybrids and offspring can produce viable seeds. The seven different characteristics of peas he selected were round or wrinkled ripe seeds, yellow or green endosperm, grey-brown or white seed coats,smooth or wrinkled seed pods, green or yellow unripe pods, axial flowers along the stems or terminal flowers at the end of stems, long stems (tall) or short stems (short). He always started the experiments with true breeding plants so that their offspring would be identical to the parents and so any changes in the progeny must be due to cross breeding. In 1865 Mendel published his study of inheritance that includes three principles concerning the inheritance of traits when cross-breeding.
The principle of uniformity is that if two plants that differ in one trait are crossed the will be uniform in the chosen trait which will be either one of the parents' traits. So when parent 1 is crossed with parent 2, all the individuals of the hybrid, the first filial generation (F1) will have the same characteristic for any pair of characteristics. A unit of heredity is now called a gene and the list of different genes is called the genotype. Pairs of genes occupying the same place on different chromosomes are called alleles, e.g. the genes for round or wrinkled ripe seeds are alleles. The expression of genes in an organism is called the phenotype.
If true breeding tall plants are crossed with true breeding short plants and tall is dominant and short is recessive, the offspring, the F1 generation, will all be tall. In the following chequerboard, the two possible gametes of the tall plants are shown horizontally and the two possible gametes of the tall plants are shown vertically.
gametes
 T
 T
t
 Tt (tall)
 Tt (tall)
t
 Tt (tall)
 Tt (tall)
When an individual of the F1 generation is crossed with another individual of the same F1 generation, the resulting hybrids are members of the F2 generation. The principle of segregation is that the individuals of the F2 generation are not uniform. Hereditary traits occur in pairs, one of each pair being inherited from each parent. During meiosis when gametes are formed, each pair of alleles, e.g. tall seeds / short seeds in the parent cells segregate (separate) into different gametes. So any gamete carries either the tall seeds or short seeds gene but not both and not neither.
In the following chequerboard, the gametes produced by the two parents, shown horizontally and vertically, are either tall or short. If tall is dominant and short is recessive, let T = tall and t = short. The genotype of the offspring can be TT (tall) or tT (tall) or Tt (tall) or tt (short) so the ratio of phenotypes is 3 tall plants to 1 short plant. This study of one allele is called a monohybrid cross
gametes
 T
 t
T
 T T (tall)
 t T (tall)
t
 T t (tall)
 t t (short)
Similarly the allele round or wrinkled seeds, round is dominant and wrinkled is recessive. Let R = round and r = wrinkled. The result of the monohybrid cross would be a ratio of 3 (round) to 1 (wrinkled) seeds. The genotypes TT and tt are said to be homozygous. The genotype Tt (or tT) is said to be heterozygous.
The principle of independent assortment is that each trait is inherited independently of the other traits so new combinations of traits can occur which were not existing before. The segregation of one pair of alleles is independent of of the segregation of any other pair of alleles. This principle is valid only for genes on different chromosomes.
For a dihybrid cross the offspring have 16 possible different genotypes and the ratio of phenotypes is 9 (tall round) 3 (tall wrinkled) 3 (short round) 1 (short wrinkled).
gametes
 TR
 Tr
 tR
 tr
TR
 TRTR (tall round)
 TrTR (tall round) tRTR (tall round) trTR (tall
round)
Tr
 TRTr (tall round) TrTr (tall wrinkled)
 tRTr (tall round) trTr (tall wrinkled)
tR
 TRtR (tall round) TrtR (tall
round)		 tRtR (short round)
 trtR (short round)
tr
 TRtr (tall round) Trtr (tall wrinkled) tRtr (short round) trtr (short wrinkled)
Later, geneticists interpreted Mendel's principles as "Mendel's three laws" while others referred to principle 2 and 3 as Mendel's first and second law.
9.6.4.1 Mendel's first law, law of uniformity (inheritance of one pair of characteristics)
According to Mendel's first law when homozygous strains are crossed which differ by one or more characteristics, the offspring in the first filial generation (F1) will all have the same characteristics. This is called the law of uniformity. To investigate the validity of Mendel's first law, put 8 unfertilized female fruit flies, normal wild strain (+) and 15 males, ebony strain (e) in each of two breeding flasks containing Drosophila medium. Then do a reciprocal hybridization with 8 unfertilized females, ebony strain (e) and 5 males, normal wild strain (+) in each of two breeding flasks containing the Drosophila medium. Leave the four breeding flasks to stand at room temperature. Ten days after the first fruit flies have hatched, note the body colour of the first filial generation (F1) offspring in the four breeding flasks. Count the numbers of each sex after anaesthetizing them with diethyl ether. Use a magnifying glass to distinguish between males and females. Compare the body colour of the first filial generation (F1) fruit flies with the body colour of the parents. Note the sex ratio. The parent fruit flies are homozygous for the characteristics under investigation. The genotypes are +/+ for fruit flies with normal body colour and e/e for fruit flies with dark body colour. So the germ cells are + or + and e or e. Insert all possible combinations of the genes in the chequerboard diagram below. Note whether the theoretical result agrees with the practical result of this experiment.
. (+) male (+) male
(e) female .
 .
(e) female . .
. (e) male (e) male
(+) female . .
(+) female . .
9.6.4.2 Mendel's first law, law of uniformity (inheritance of two pairs of characteristics)
Put 8 unfertilized female fruit flies, ebony strain (e) and 15 male fruit flies, vestigial strain (vg) in each of two breeding flasks containing the Drosophila medium. Leave the breeding flasks to stand at room temperature. Ten days after the first fruit flies have hatched, note the body colour and wing shape of the first filial generation (F1) offspring in both breeding flasks. Anaesthetize the fruit flies with diethyl ether. Place them on a sheet of filter paper and examine them under a magnifying glass. Compare their appearance with the parents The parent fruit flies, which were crossbreeds, possessed the following observable characteristics:
.
 Body colour Wing shape
Male Normal wild (+) Stump-winged, vestigial (vg)
Female Dark, ebony (e) Normal -wild (+)
The genotypes are the following: male vg / vg // + / +, female, + / + // e / e. To find what types of germ cells can be formed, insert all possible combinations of the genes in the chequerboard. Note whether the theoretical results agree with the experimental results. Draw a chequerboard and predict whether the result of the experiment would have been the same if both recessive characteristics were introduced by one parent in the hybridization.
.
 male (+) male (vg)
female (e) (+) (e) (vg) (e)
female (+) (+) (+) (vg) (+)
9.6.5 Mendel's second law, law of segregation
According to Mendel's second law, the characteristics of the parent generation recur in the second filial generation (F2) in a quite specific numerical ratio. Investigate the segregation of the dominant recessive characteristic pair normal wild/ebony (+/e) of the fly in the second filial generation. Put 8 female and 5 male fruit flies from one of the first filial generations (F1) from the previous experiment in each of two breeding flasks containing Drosophila medium. Leave the breeding flasks to stand at room temperature. Ten days after the first fruit flies have hatched, note the body colour of the second filial generation (F2) offspring. Anaesthetize the fruit flies with diethyl ether and place them on a sheet of filter paper. Note the numerical ratio of the body colours of the parent fruit flies, Normal wild (+) and ebony (e). The genotype of the fruit flies of the first filial generation (F1) crossed in this experiment is +/e. The germ cells they can form are (+) or (e). Insert all possible combinations of the genes in a checker-board. Does the theoretical result obtained with the checker-board agree with the practical result from the hybridization experiment? Make a comparative table using both sets of results. Calculated numerical values from the checker-board and the numbers obtained by counting in the experiment. Note the deviations. Note whether all fruit flies having the same appearance have the same genetic constitution, i.e. do fruit flies with the same phenotype always have the same genotype.
. male male
female .
 .
female . .
9.6.6 Mendel's third law, law of independent assortment
According to Mendel's second Law the characteristics of the parent generation recur in an exact numerical ratio (subdivision ratio) In the second filial generation (F2) following the Law of Segregation. If different pairs of characteristics (alleles) are not contained on the same chromosome, they are distributed (assorted) independently during the formation of the germ cells and freely recombine. This process is called Mendel's third Law, the Law of Independent Assortment. Investigate the free recombination of genes, Mendel's third Law, in the fruit fly. Place 8 female and 15 male fruit flies from the first filial (F1) generation of the previous experiment in each of two breeding flasks containing Drosophila medium. Leave the breeding flasks to stand at room temperature. About 10 days after the first fruit flies have hatched, examine the offspring in both breeding flasks, the second filial generation (F2) for their body colour and wing shape. For this purpose, anaesthetize the fruit flies with diethyl ether. Put them on a sheet of filter paper and examine them under a magnifying glass. Note the number of different characteristic types, phenotypes. Count each of the different types. Note what phenotype has occurred for the first time. In the previous experiment, you crossed stump winged (vestigial, vg) and dark bodied (ebony, e) fruit flies of the parent generation. Their genotypes were vg/vs and +/+ respectively. +/+ e/e The offspring, in accordance with their genotypes +/vg, are normal coloured and have normal shaped wings. e/+. These fruit flies were crossed in the present experiment. Note what type of germ cells they can they. Insert all possible combinations of the genes in a checker-board. Determine the external characteristics of the fruit flies according to their respective genes. Note which characteristic types (phenotypes) must occur and how often. Note whether this theoretical result agrees with the practical result of the experiment.
. male male male male
female . . . .
female . . . .
female . . . .
female . . . .
9.6.7 Sex.linked inheritance, introduction of a recessive characteristic by female
The female has the pair of chromosomes (X/X) but he male has the genotype (X/Y) so the genes on the X chromosome are distributed differently in hybridization from those on the other chromosomes.
Investigate the inheritance of the sex.linked recessive characteristic white (white.eyed, w) in the fly introduced by the female in hybridization. Put 8 unfertilized female fruit flies, white strain (w) and 15 male, Normal.wild strain (+) in each of two breeding flasks containing Drosophila medium. Leave the breeding flasks to stand at room temperature. Ten days after the first fruit flies have hatched, note the eye colour and sex of the first filial generation (F1) offspring in both breeding flasks. Anaesthetize the fruit flies with diethyl ether. Put them on a sheet of filter paper and examine them under a magnifying glass. Record the results in a table as follows:
. male female
White eyes . .
Normal.coloured
(red) eyes		 . .
Note whether the result in accordance with Mendel's first Law. The gene for eye colour is on the X chromosome. The females have two X chromosomes. In this experiment they are white.eyed and their genotype is w/w. The males possess only one X chromosome. In this experiment they have normal coloured (red) eyes and their genotype is +/y. Note what type of germ cells can be formed and insert all possible combinations of the genes in a checker.board that follows. Determine the external characteristics and sex of the files according to their respective genes. Note whether this theoretical result agrees with the result of the experiment.
. male male
female . .
female . .
2. Put 8 females and 5 males of the first filial generation (F1) from the above experiment in each of two breeding flasks containing Drosophila medium. Leave the breeding flasks to stand at room temperature. Ten days after the first fruit flies have hatched, note the eye colour and sex of the second filial generation (F2) offspring in both breeding vessels. Anaesthetize the fruit flies with diethyl ether, place them on a sheet of filter paper and examine them under a magnifying glass. Count the files and record the results in the following table:.
. male female
white eyes . .
normal coloured (red) eyes . .
Note the genotype of both the males and females of the first filial generation. Note the types of germ cells they can be formed. Insert all possible combinations of genes in the checker.board that follows.
Determine the external characteristics, phenotypes and sex of the fruit flies according to their respective genes. Note whether the theoretical results agree with the practical result of the experiment.
. male male
female . .
female . .
9.6.8 Sex.linked inheritance, introduction of a recessive characteristic by a male
1. In the preceding experiments you investigated the heredity process when a sex.linked recessive characteristic is introduced by the female during hybridization. Investigate the heredity mechanism when this characteristic is introduced by the male when the sex.linked recessive characteristic white (white eyes, w) in the fly is introduced by the male. Put 8 unfertilized female fruit flies, Normal.wild strain (+) and 15 males, white strain (w) in each of two breeding flasks containing Drosophila medium. Leave the breeding flasks to stand at room temperature. Ten days after the first fruit flies have hatched, note the colour of the eyes of the first filial generation (F1) offspring in both breeding flasks. Anaesthetize the fruit flies with diethyl ether. Put them on a sheet of filter paper and examine them under a magnifying glass. Note whether the result is in accordance with Mendel's first Law. Note the genotypes of the parent fruit flies and the germ cells can they form. Insert all possible combinations of the genes in the following chequerboard.
Determine the external characteristics, phenotypes, of the fruit flies according to their respective genes. Note whether the theoretical result obtained with the chequerboard agrees with the practical result of the experiment.
. male male
female . .
female . .
2. Put 8 females and 15 males of the first filial generation (F1) from the above experiment in each of two breeding flasks containing Drosophila medium. Leave the breeding flasks to stand at room temperature. Ten days after the first fruit flies have hatched, note the eye colour and sex of the second filial generation (F2) offspring in both breeding flasks. Anaesthetize the fruit flies with diethyl ether. Put them on a sheet of filter paper and examine them under a magnifying glass. Count the fruit flies and record the results in the following table.
. male female
white eyes . .
normal coloured (red) eyes . .
Note the ratio of red.eyed to white.eyed fruit flies. Note how the eye colour is distributed between the sexes and the genotype of the fruit flies of the first filial generation. Note what types of germ cells these fruit flies can form. Insert all possible combinations of the genes in the following chequerboard. Determine the external characteristics (phenotypes) and sex of the fruit flies according to their respective genes. Note whether this theoretical result agrees with the practical result of the hybridization experiment.
. male male
female . .
female . .
9.6.9 Genetics, lethal factors
After many mutations, the genotype may become altered to such an extent that the offspring are no longer viable. Genes that mutate in this way are called lethal factors, e.g. (CY) mutant of the fruit fly. These fruit flies have upwards curving wings. The characteristic is dominant. Fertilized egg cells in which (CY) it is homozygous do not develop. Put 8 unfertilized female and 15 male fruit flies, curly strain (Cy) in each of two breeding flasks containing Drosophila medium. Leave the breeding flasks to stand at room temperature. Ten days after the first fruit flies have hatched, note the shape of the wings of the offspring in each breeding container. Anaesthetize the fruit flies with diethyl ether. Put them on a sheet of filter paper and examine them under a magnifying glass. Count the fruit flies with upwards curving wings and normal wings. The dominant characteristic Curly homozygous fruit flies, genotype (CYCY) cannot exist so the hybrid fruit flies must have the genotype Cy/+. Insert all possible combinations of the genes in the chequerboard. Determine from the checker.board what ratio there ought to be of fruit flies with upwards curving to normal wings in the first filial generation (F1). Note whether this theoretical value corresponds to the practical result of the experiment. Note the apparent deviation from Mendel's Laws and why it occurs.
. male male
female . .
female . .
9.6.10 Giant chromosomes
Investigate the form and structure of the giant chromosomes in the salivary gland cells of the larva of the fruit fly. Press carefully on the coverslip with the handle stem of a dissecting needle until the salivary glands disintegrate into individual cells. Apply slightly more pressure to squash the cells so that the chromosomes are pressed out. Use filter paper to draw off the acetic acid which has spread under the coverslip and examine the preparation under a microscope with preferably an oil immersion lens, magnification not less than 40 x. Put a larva of the fly in a large drop of carmine acetic acid on a microscope slide. Hold down the larva by pressing the side of a dissecting needle horizontally across it, about one third of the way along from the end of the abdomen. Pierce the exoskeleton by pressing the point of another dissecting needle at a slight angle from the horizontal into the larva between the second and third segment, i.e. directly behind the fauces. Pull the head section of the larva forwards until it becomes detached from the rest of the body, pulling the organs attached to it out of the body. Identify the salivary glands from by their club.like shape and glazed appearance. Remove the attached fatty tissue. Transfer the salivary glands to a fresh drop of carmine acetic acid on a second slide and place a coverslip over them. After 2 minutes place a drop of 45% acetic acid on the edge of the coverslip and draw it across under the coverslip by placing the edge of a filter paper strip at the opposite edge of the glass. The chromosomes in the salivary gland cells of the larvae of fruit flies are so large they are called giant chromosomes. These chromosomes form by repeated longitudinal division of the chromatic thread without subsequent splitting of the products of the division. Note their shape and structure.