1st Semester 57 - BIOLOGY JUNCTION

Genetics of Drosophila Melanogaster

Genetics of Drosophila melanogaster

Introduction:
Gregor Mendel revolutionized the study of genetics. By studying genetic inheritance in pea plants, Gregor Mendel established two basic laws of that serve as the cornerstones of modern genetics: Mendel’s Law of Segregation and Law of Independent Assortment. Mendel’s Law of Segregation says that each trait has two alleles, and that each gamete contains one and only one of these alleles. These alleles are a source of genetic variability among offspring. Mendel’s Law of Independent Assortment says that the alleles for one trait separate independently of the alleles for another trait. This also helps ensure genetic variability among offspring.
Mendel’s laws have their limitations. For example, if two genes are on the same chromosome, the assortment of their alleles will not be independent. Also, for genes found on the X chromosome, expression of the trait can be linked to the sex of the offspring. Our knowledge of genetics and the tools we use in its study have advanced a great deal since Mendel’s time, but his basic concepts still stand true.
Drosophila melanogaster, the common fruit fly, has been used for genetic experiments since T.H. Morgan started his experiments in1907. Drosophila make good genetic specimens because they are small, produce many offspring, have easily discernable mutations, have only four pairs of chromosomes, and complete their entire life cycle in about 12 days. They also have very simple food requirements. Chromosomes 1 (the X chromosome), 2, and 3 are very large, and the Y chromosome – number 4 – is extremely small. These four chromosomes have thousands of genes, many of which can be found in most eukaryotes, including humans.
Drosophila embryos develop in the egg membrane. The egg hatches and produces a larva that feeds by burrowing through the medium. The larval period consists of three stages, or instars, the end of each stage marked by a molt. Near the end of the larval period, the third instar will crawl up the side of the vial, attach themselves to a dry surface, and form a pupae. After a while the adults emerge.
Differences in body features help distinguish between male and female flies. Females are slightly larger and have a light-colored, pointed abdomen. The abdomen of males will be dark and blunt. The male flies also have dark bristles, sex combs, on the upper portion of the forelegs.

Hypothesis:
After performing a dihybrid cross between males with normal wings and sepia eyes and females with vestigial wings and red eyes, we expect to see only hybrids with normal wings and red eyes in the first filial generation. Then we expect to observe a 9:3:3:1 ratio of phenotypes in the second filial generation.

Materials and Methods:
The materials used for this lab were: culture vial of dihybrid cross, isopropyl alcohol 10%, camel’s hair brush, thermo-anesthetizer, petri dish, 2 Drosophila vials and labels, Drosophila medium, fly morgue.

A vial of wild-type Drosophila was thermally immobilized and the flies were placed in a petri dish. Traits were observed. A vial of prepared Drosophila was immobilized and then observed under a dissecting microscope. Males and females were separated and mutations were observed and recorded. The parental generation was placed in the morgue. The vial was placed in an incubator to allow the F1 generation to mature.
The F1 generation was immobilized and examined under a dissecting microscope. The sex and mutations of each fly were recorded. Five mating pairs of the F1 generation were placed into a fresh culture vial, and the vial was placed in an incubator. The remaining F1 flies were placed in the morgue. The F1 flies were left in the vial for about a week to mate and lay eggs. Then the adults were removed and placed in the morgue. The vial was placed back in the incubator to allow the F2 generation to mature. The F2 generation was immobilized and examined under a dissecting microscope. The sex and mutations of each fly were recorded.

Results:

Table 1 Phenotypes of the Parental Generation

Phenotypes	Number of Males	Number of Females
Normal wings/red eyes	0	0
Normal wings/sepia eyes	3	0
vestigial wings/red eyes	0	4
vestigial wings/sepia eyes	0	0

Table 2 Phenotypes of the F1 Generation

Phenotype	Number of Males	Number of Females
Normal wings/red eyes	78	95
Normal wings/sepia eyes	0	0
vestigial wings/red eyes	0	0
vestigial wings/sepia eyes	0	0

Table 3 Phenotypes of the F2 Generation

Phenotypes	Number of Males	Number of Females
Normal wings/red eyes	4	7
Normal wings/sepia eyes	4	5
vestigial wings/red eyes	0	1
vestigial wings/sepia eyes	0	0
normal red/mutated body shape	2	0
normal sepia/mutated body shape	1	0

Questions

How are the alleles for genes on different chromosomes distributed to gametes? What genetic principle does this illustrate?
The alleles on different chromosomes are distributed independently of one another, demonstrating Mendel’s Law of Independent Assortment.
Why was it important to have virgin females for the first cross (yielding the F1 generation), but not the second cross (yielding the F2 generation)?
It was important to have virgin females for the first cross to ensure that the offspring are the result of the desired cross. It was not necessary to isolate virgin females for the second cross because the only male flies to which they had been exposed were also members of the F1 generation.
What did the chi-square test tell you about the validity of your experiment data? What is the importance of such a test?
The chi-square test showed that the results of our first cross were valid, but that the results of our F1 cross were not normal. It is important to conduct such a test to determine how much your experimental data deviated from what was expected.

Discussion and Conclusion:
The results of our parental cross turned out just as expected, but our F2 generation was not normal. Some sort of mutation must have occurred that caused the strange body shape seen in several individuals of our F2 generation.

BACK

Genetics Problems ppt Questions

Genetics Problems
ppt Questions

Independent Assortment

1. How many different kinds of gametes could the following individuals produce? Remember the formula 2n where n equals the number of heterozygotes.

a. aaBb

b. CCDdee

c. AABbCcDD

d. MmNnOoPpQq

e. UUVVWWXXYYZz

P1, F1, and F2 Monohybrid Crosses

2. In dogs, wire-haired is due to a dominant gene (W), smooth-haired is due to its recessive allele (w). Show the results of crossing a homozygous wire-haired dog with a smooth-haired dog.

3. What kind of cross is this?

4. What was the genotype of all of the puppies? the phenotype?

5. The puppies belong to the _________ generation.

6. How would you write the F1 cross for this trait?

7. Show the results of working the F1 cross for this trait.

6. What phenotypic ratio did you get from this F1 cross?

7. What genotypic ratio did you get from this F1 cross?

8. Two wire-haired dogs are mated. Among the offspring of their first litter is a smooth-haired pup. If these two dogs mate again, what are the chances of them having another smooth-haired pup?

9. What are the chances that the pup will be wire-haired?

10. A Wire-haired male is mated with a smooth-haired female. The mother of the wire-haired male was smooth-haired. What are the phenotypes and genotypes of the pups they could produce? Show how you got your results.

Incomplete Dominance

11. In snapdragons, red flower color (R) is incompletely dominant over white flower color (r). The hybrids or heterozygous plants (Rr) are pink in color. Show the genotype for a white flower and for a red flower.

12. If a red-flowered plant is crossed with a white-flowered plant, what are the genotypes and phenotypes of the F1 generation plants? Show your work.

13. What is the phenotype of the flowers? what is their genotype?

14. What genotypes and phenotypes will be produced in the F2 generation? Show your work.

15. How did the genotypic and phenotypic ratio compare to each other in this incomplete dominance cross?

16. What would the phenotypic ratio have been if this had been complete dominance?

17. What kind of offspring can be produced if a red-flowered plant is crossed with a pink-flowered plant? Show your work.

18. What kind of offspring is/are produced if a pink-flowered plant is crossed with a white-flowered plant? Show your work.

Sex-linked Traits

19. What is the genotype for female? for male?

20. In humans, colorblindness (Xc) is a recessive sex-linked trait. Two people with normal color vision (XC) have a colorblind son. What are the genotypes of the parents?

21. What are the genotypes and phenotypes possible among their other children? Show your work.

22. A couple has a colorblind daughter. What are the possible genotypes and phenotypes of the parents and the daughter?

Dihybrid Crosses

23. In humans, the presence of freckles is due to a dominant gene (F) and the non-freckled condition is due to its recessive allele (f). Dimpled cheeks (D) are dominant to non-dimpled cheeks (d). Two persons with freckles and dimpled cheeks have two children. One child has freckles but no dimples. The other child has dimples but no freckles. What is the genotypes of the parents? the children?

24. What are the possible phenotypes and genotypes of the children that they could produce? Show all your work.

25. What phenotypic ratio did you get?

26. What genotypic ratio did you get?

27. What are the chances that they would have a child whom lacks both freckles and dimples? What would be the child’s genotype?

28. A person with freckles and dimples whose mother lacked both freckles and dimples marries a person with freckles but no dimples whose father did not have freckles or dimples. What are the chances that they would have a child whom lacks both freckles and dimples? Show the genotypes of the parents and all the offspring.

29. In dogs, the inheritance of hair color involves a gene (B) for black hair and a gene (b) for brown hair. A dominant (C) is also involved. It must be present for the color to be synthesized (made). If this gene is NOT present, a blond condition results. Complete the following table:

Genotype	Phenotype	Color Deposition gene
BB or Bb		CC or Cc
bb		CC or Cc
BB or Bb		cc
bb		cc

30. A brown-haired male, whose father was a blond, is mated with a black-haired female ,whose mother was brown-haired and her father was blond. What is the genotype of the man and woman? Show the genotypes and phenotypes of all of their offspring.

Population Genetics or Hardy-Weinberg Law

Sixteen percent (16%) of the human population is known to be able to wiggle their ears. This trait is determined to be a recessive gene. Use the following equations to answer this population genetics problem:

1 = p2 + 2pq + q2 then use 1 = p + q

p2 – frequency of homozygous dominants

2pq – frequency of heterozygotes

q2 – frequency of homozygous recessives

p – frequency of dominant allele

q – frequency of recessive allele

31. What percent of the population is homozygous dominant for this trait? Show your work.

32. What percent of the population is heterozygous for this trait? Show your work.

Multiple Alleles – ABO Blood Type

33. Henry Anonymous, a film star, was involved in a paternity case. The woman bringing the suit had two children. One child had blood type A and the other child had blood type B. Her blood type was O, the same as Henry’s. The judge in the case awarded damages to the woman, saying that Henry had to be the father of at least one of her children. was the judge correct in his decision? Show how you got your answer.

Extracting DNA

Extract DNA from Anything Living

Introduction:

Since DNA is the blueprint for life, everything living contains DNA. DNA isolation is one of the most basic and essential techniques in the study of DNA. The extraction of DNA from cells and its purification are of primary importance to the field of biotechnology and forensics. Extraction and purification of DNA are the first steps in the analysis and manipulation of DNA that allow scientists to detect genetic disorders, produce DNA fingerprints of individuals, and even create genetically engineered organisms that can produce beneficial products such as insulin, antibiotics, and hormones.

DNA can be extracted from many types of cells. The first step is to lyse or break open the cell. This can be done by grinding a piece of tissue in a blender. After the cells have broken open, a salt solution such as NaCl and a detergent solution containing the compound SDS (sodiumdodecyl sulfate) is added. These solutions break down and emulsify the fat & proteins that make up a cell membrane. Finally, ethanol is added because DNA is soluble in water. The alcohol causes DNA to precipitate, or settle out of the solution, leaving behind all the cellular components that aren’t soluble in alcohol. The DNA can be spooled (wound) on a stirring rod and pulled from the solution at this point.

Just follow these 3 easy steps:

Detergent, eNzymes (meat tenderizer), Alcohol

Objective:

To extract DNA from cells.

Materials:

Blender, split peas, salt, detergent, water, measuring cup and spoons, strainer, meat tenderizer, alcohol, test tube, glass stirring rod

Procedure:

First, you need to find something that contains DNA such as split peas, fresh spinach, chicken liver, onion, or broccoli.

Measure about 100 ml or 1/2 cup of split peas and place them in a blender.
Add a large pinch of salt (less than 1 ml or about 1/8 teaspoon) to the blender.
Add about twice as much cold water as the DNA source (about 200 ml or 1 cup) to the peas in the blender.
Blend on high (lid on) for about 15 seconds.

The blender separates the pea cells from each other, so you now have a really thin pea-cell soup.

And now, those 3 easy steps:

Pour your thin pea-cell soup through a strainer into another container like a measuring cup or beaker.

Estimate how much pea soup you have and add about 1/6 of that amount of liquid detergent (about 30ml or 2 tablespoons). Swirl to mix.

Let the mixture sit for 5-10 minutes.

The detergent captures the proteins & lipids of the cell membrane.

Pour the mixture into test tubes or other small glass containers, each about 1/3 full.
Add a pinch of enzymes to each test tube and stir gently. Be careful! If you stir too hard, you’ll break up the DNA, making it harder to see. (Use meat tenderizer for enzymes. If you can’t find tenderizer, try using pineapple juice or contact lens cleaning solution.)

The DNA in the nucleus of the cell is molded, folded, and protected by proteins. The meat tenderizer cuts the proteins away from the DNA.

Tilt your test tube and slowly pour rubbing alcohol (70-95% isopropyl or ethyl alcohol) into the tube down the side so that it forms a layer on top of the pea mixture. Pour until you have about the same amount of alcohol in the tube as pea mixture.

Alcohol is less dense than water, so it floats on top forming two separate layers.
All of the grease and the protein that we broke up in the first two steps move to the bottom, watery layer.
DNA will rise into the alcohol layer from the pea layer. You can use a glass stirring rod or a wooden stick to draw the DNA into the alcohol.
Slowly turning the stirring rod will spool (wrap) the DNA around the rod so it can be removed from the liquid.

Questions:

1. Does the DNA have any color?

2. Describe the appearance of the DNA.

3. Do only living things contain DNA? Explain.

Frequently Asked Questions: 1. I’m pretty sure I’m not seeing DNA. What did I do wrong?

First, check one more time for DNA. Look very closely at the alcohol layer for tiny bubbles. Often, clumps of DNA are loosely attached to the bubbles.

If you are sure you don’t see DNA, then the next step is to make sure that you started with enough DNA in the first place. Many food sources of DNA, such as grapes, also contain a lot of water. If the blended cell soup is too watery, there won’t be enough DNA to see. To fix this, go back to the first step and add less water. The cell soup should be opaque, meaning that you can’t see through it. Another possible reason for not seeing any DNA is not allowing enough time for each step to complete. Make sure to stir in the detergent for at least five minutes. If the cell and nuclear membranes are still intact, the DNA will be stuck in the bottom layer. Often, if you let the test tube of pea mixture and alcohol sit for 30-60 minutes, DNA will precipitate into the alcohol layer.

2. Why does the DNA clump together?

Single molecules of DNA are long and stringy. Each cell of your body contains six feet of DNA, but it’s only one-millionth of an inch wide. To fit all of this DNA into your cells, it needs to be packed efficiently. To solve this problem, DNA twists tightly and clumps together inside cells. Even when you extract DNA from cells, it still clumps together, though not as much as it would inside the cell.

Imagine this: the human body contains about 100 trillion cells, each of which contains six feet of DNA. If you do the math, you’ll find that our bodies contain more than a billion miles of DNA!

3. Can I use this DNA as a sample for gel electrophoresis?

Yes, but all you will see is a smear. The DNA you have extracted is genomic, meaning that you have the entire collection of DNA from each cell. Unless you cut the DNA with restriction enzymes, it is too long and stringy to move through the pores of the gel; instead, all you will end up seeing is a smear.

4. Isn’t the white, stringy stuff actually a mix of DNA and RNA?

That’s exactly right! The procedure for DNA extraction is really a procedure for nucleic acid extraction. However, much of the RNA is cut by ribonucleases (enzymes that cut RNA) that are released when the cells are broken open.

Fermentation Rootbeer

FERMENTATION – MAKING ROOT BEER
David Fankhauser’s Main Page

Introduction:

Fermentation has been used by mankind for thousands of years for raising bread, fermenting wine and brewing beer. The products of the fermentation of sugar by baker’s yeast Saccharomyces cerevisiae (a fungus) are ethyl alcohol and carbon dioxide. Carbon dioxide causes bread to rise and gives effervescent drinks their bubbles. This action of yeast on sugar is used to ‘carbonate’ beverages, as in the addition of bubbles to champagne).

We will set up a fermentation in a closed system and capture the generated carbon dioxide to carbonate root beer. You may of course adjust the quantities of sugar and/or extract (Zatarain’s) to taste.

	EQUIPMENT	SUPPLIES
	clean 2 liter plastic soft drink bottle with cap funnel 1 cup measuring cup 1/4 tsp measuring spoon 1 Tbl measuring spoon	Cane (table) sugar [sucrose] (1 cup) Zatarain’s Root Beer Extract (1 tablespoon) (When I could not find it locally, I ordered a case of 12 bottles for $18 from Zatarain’s, New Orleans, LA 70114 powdered baker’s yeast (1/4 teaspoon) (Yeast for brewing would certainly work at least as well as baking yeast.) cold fresh water

INSTRUCTIONS:

	1) Assemble the necessary equipment and supplies
	2) With a dry funnel, add in sequence: 1 level cup of table sugar (cane sugar) (You can adjust the amount to achieve the desired sweetness.)
	3) Add: 1/4 teaspoon powdered baker’s yeast ( fresh and active) (Fleischmann’s or other brand)
	4) You can see the yeast granules on top of the sugar.
	5) Shake to distribute the yeast grains into the sugar.
	6) Swirl the sugar/yeast mixture in the bottom to make it concave (to catch the extract).
	7) Add with funnel: 1 Tbl of root beer extract (I prefer Zatarain’s, but Hires, etc. will work.) on top of the dry sugar
	8) The extract sticks to the sugar which will help dissolve the extract in the next steps.

9) Half fill the bottle with fresh cool tap water (the less chlorine, the better).

Rinse in the extract which sticks to the tablespoon and funnel. Swirl to dissolve the ingredients.

10) Q.s. [fill up] to the neck of the bottle with fresh cool tap water, leaving about an inch of head space, securely screw cap down to seal. Invert repeatedly to thoroughly dissolve.

If you leave it in a warm temperature longer than two weeks, you risk an explosion…

11) Place at room temperature about three to four days until the bottle feels hard to a forceful squeeze. Move to a cool place (below 65 F). refrigerate overnight to thoroughly chill before serving. Crack the lid of the thoroughly chilled root beer just a little to release the pressure slowly.

NOTE: Do not leave the finished root beer in a warm place once the bottle feels hard. After a couple weeks or so at room temperature, especially in the summer when the temperature is high, enough pressure may build up to explode the bottle! There is no danger of this if the finished root beer is refrigerated.

12) Move to a refrigerator overnight before opening.

NOTE: There will be a sediment of yeast at the bottom of the bottle, so that the last bit of root beer will be turbid. Decant carefully if you wish to avoid this sediment.

A WORD ABOUT THE ALCOHOL IN HOME MADE ROOT BEER: The alcoholic content which results from the fermentation of this root beer and found it to be between 0.35 and 0.5 %. Comparing this to the 6% in many beers, it would require a person to drink about a gallon and a half of this root beer to be equivalent to one 12 ounce beer. I would call this amount of alcohol negligible, but for persons with metabolic problems who cannot metabolize alcohol properly, or religious prohibition against any alcohol, consumption should be limited or avoided.

Food Chemistry

Introduction:

All living things are made up of one or more cells, and the cells in turn contain many kinds of molecules. In this lab we will be looking at several different macromolecules (large molecules): proteins, carbohydrates, and lipids (specifically fats). Various chemicals will be used in this laboratory to test for the presence of these molecules. Most often, you will be looking for a particular color change. If the change is observed, the test is said to be positive because it indicates that a particular molecule is present. If the color change is not observed, the test is said to be negative because it indicates that a particular molecule is not present.

You will be using these tests to determine which of the macromolecules are present in various samples of food.

In all of the procedures, you will need to include a distilled water sample as a control. Usually, a control goes through all the steps of the experiment but lacks one essential factor (the experimental variable). This missing factor allows you to observe the difference between a positive result and a negative result. If the control sample tests positive, you know your test is invalid. Some tests may also contain other controls to make sure certain additives are not contaminated with the substance for which you are testing.

Proteins:

Protein molecules are long chains of amino acids joined by peptide bonds.

Biuret reagent, which is a blue color, contains a strong solution of sodium or potassium hydroxide (NaOH or KOH) and a very small amount of very dilute copper sulfate (CuSO4) solution. The reagent changes color in the presence of proteins or peptides because the amino group (H2N-) of the protein or peptide chemically combines with the copper ions in biuret reagent.

Carbohydrates:

Carbohydrates include sugars and molecules that are chains of sugars. Glucose is a simple sugar, also known as a monosaccharide. Sucrose, or table sugar is a disaccharide, two sugar units bonded together. Starch is a polysaccharide, a long chain of glucose units.

Benedict’s reagent reacts with many sugars (both mono- and disaccharides) after being heated in a boiling water bath. Increasing concentrations of sugar give a continuum of colored products ranging from green through yellow and orange to orange red.

Iodine solution reacts with starch to produce colors ranging from a brownish to blue black.

Lipids:

Lipids are hydrophobic molecules which are insoluble in water and soluble in solvents such as alcohol and ether. Lipids include fats, oils, and cholesterol.

Lipids do not evaporate from brown paper, instead leaving an oily spot. Lipids also do not mix with water, forming a separate layer, usually on top of the water. However, some molecules mix with both water and lipids, and can be used to mix the two. These molecules are known as emulsifiers. The liver produces bile salts which act as emulsifiers in the digestive tract. Soaps and detergents also act as emulsifiers.

Summary of tests:

Biuret Reagent	Benedict’s Reagent	Iodine Solution	Brown Paper
Reacts with proteins or peptides, turns purple (protein) or pink (peptides)	Reacts with sugars, turns green through yellow to orange (green, less sugar, orange, more sugar)	Reacts with starch, turns dark brown to black	Lipids leave oily spot

Procedure:

Take some time to plan with your lab partner what tests you will do, and in what order before beginning the procedures.

There are available in the lab a variety of different types of common food. Choose at least 3 foods and test each for the presence of protein, carbohydrate (both starch and simple sugars), and fats. Be sure to plan your experiments before starting.

Form a hypothesis for each sample you have chosen to test.

Samples will need to be suspended in water for most tests. Using a mortar and pestle if necessary, break each sample to be tested into small pieces and suspend the pieces in a small amount of distilled water.

Also available are samples of each of the types of molecules for which you will be testing. Use these samples to try out the tests so that you will know what a positive result looks like.

Be sure to include a blank control (distilled water) with each test so you know what a negative result looks like. You may also include a positive control, a sample which you know contains the substance for which you are testing.

The procedures for testing for each type of molecule are given below.

Proteins and Peptides

Proteins:

Use a separate test tube for each sample to be tested, as well as one for a control.
Label each test tube.
Place about 1 mL of each sample (and control) in its test tube.
Add 5 drops of copper sulfate solution to each tube.
Add 10 drops of potassium hydroxide solution to each tube and mix.
Record the tube contents and final color in a data table.
Conclusions: which tubes contained protein?

Carbohydrates: Sugars and starch

Starch

Use a separate test tube for each sample to be tested, as well as one (or two) for a control.
Label each test tube.
Place about 1 mL of each sample (and control) in its test tube.
Add 5 drops of iodine solution to each tube and mix.
Record the tube contents and final color in a data table.
Conclusions: which tubes contained starch?

Sugar

Use a separate test tube for each sample to be tested, as well as one (or two) for a control.
Label each test tube.
Place about 1 mL of each sample (and control) in its test tube.
Add about 2 mL of Benedict’s reagent to each tube and mix.
Heat the tubes in a boiling water bath for 5-10 minutes.
Record the tube contents and final color in a data table.
Conclusions: which tubes contained sugar?

Lipids

Place a small sample of the material to be tested on a square of brown paper.
Place a small drop of water on the square of brown paper.
Compare the drop of water to the sample.
Wait at least 5 minutes. Evaluate which substance impregnates the paper and which is subject to evaporation. Record your results.
Conclusions: which sample contained lipids?

Conclusion Questions:

Why do experimental procedures include control samples?
How would you test an unknown solution for each of the following:
1. Sugars
2. Fat
3. Starch
4. Protein
Assume that you have tested an unknown sample with both biuret solution and Benedict’s solution and that both tests result in a blue color. What have you learned?
What purpose is served when a test is done using water instead of a sample substance?
Compare your results.

Lab report:

Lab reports must include the following:

A Title to the lab. A Purpose: What was studied in this lab, and why did we study it?
Procedure: a brief description of each type of test, what constitutes a positive test and what constitutes a negative test.
All data tables.
For each food sample, state your hypothesis and your conclusions. Did your results confirm or refute your hypothesis?
Answers to questions.
A brief analysis of what worked in this lab and what didn’t work, and why.