Unit 4 Genetics) Labs

Investigation

Probability*

The probability of a chance event can be calculated mathematically using the following formula: Probability = number of events of choice/number of possible events What is the probability that you will draw a spade from a shuffled deck of cards? There are 52 cards in the deck (52 possible events). Of these 13 cards are spades (13 events of choice). Therefore, the probability of drawing a spade from this deck is 13/52 (or ¼ or .25 or 25%). To determine the probability that you will draw the ace of diamonds, you again have 52 possible events, but this time there is only one event of choice. The probability is 1/52 or 2%.In this investigation, you will determine the probability for the results of a coin toss. Materials (per team of two) 2 pennies (one shiny, one dull)

card board box

Procedure 1. Work in teams of two. One person will be student A and the other will be student B. 2. Student A will prepare a score sheet with two columns-one labeled H (heads), the other T (tails). Student B will toss a penny ten times. Toss it into the cardboard box to prevent the coin from rolling away. 3. Student A will use a slash mark (/) to indicate the result of each toss. Tally the tosses in the appropriate column on the score sheet. After 10 tosses, draw a line across the two columns and pass the sheet to student B. Student A will then make 10 tosses, and student B will tally the results. 4. Continue reversing the rolls until the results of 100 (10 series of 10) have been tallied. 5. Prepare a score sheet with four columns labeled H/H, Dull H/Shiny Tail, Dull T/ Shiny H, and T/T (H = heads; T = tails). Obtain two pennies-1 dull and 1 shiny. Toss both pennies together 20 times, while your partner tallies each result in the appropriate column of the score sheet. 6. Reverse roles once so that you have a total of 40 tosses.

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Discussion 1. How many heads are probable in a series of 10 tosses? How many did you actually observe on the first 10 tosses? 2. Deviation is a measure of the difference between the expected and observed results. It is not the difference itself. It is the ratio of the sum of the difference itself. It is the sum of the differences between expected and observed results to the total number of observations. Thus: Deviation = (| difference between heads expected and heads observed | + | difference between tails expected and tails observed |) / number of tosses Calculate the deviation for each of the 10 sets of tosses. 3. Calculate the deviation for your team’s total (100 tosses). 4. Add the data of all teams in your class. Calculate the class deviation. 5. If your school has more than one biology class, combine the data of all classes. Calculate the deviation of all classes. 6. How does increasing the number of tosses affect the average size of the deviation? These results demonstrate an important principle of probability. State what that is. 7. On the white board, record the data for tossing two pennies together. Add each column of the chart. In how many columns do data concerning heads of a dull penny appear? 8. In what fraction of the total number of tosses did heads of dull pennies occur? 9. In how columns do data concerning heads of shiny pennies occur? 10. In what fraction of the total tosses did heads of shiny pennies occur? 11. In how many columns do heads of both dull and shiny pennies appear?

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Monohybrid Crosses and the Punnett Square Name____________________________ Date_____________________________ Introduction: Scientists use a grid-like tool (Punnett Square) to make predictions about various genetic problems. The Punnett Square shows only the probability of what might occur and not the actual results. Probability is the chance of somthing occurring. If one wants to flip a coin 100 times, since there are 2 sides to the coin, he would expect 50 heads and 50 tails. If he flips the coin 100 times, he may actually get 60 heads and 40 tails. Prediction is one thing, and actually getting the predicted results is another. The Punnett square only shows the chances of what might occur each time the event is under taken. Objective: In this investigation you will use a Punnett square to predict the possible genotypes and phenotypes and their ratios from a monohybrid cross. Materials: Yellow beads Green Beads Small paper cups (one bag labeled male and the other female ) Procedure: 1. Each group of 4 students will pick up 2 paper cups filled with 15 green (G) beads and 15 yellow (g) beads. This represents 2 heterozygous parents Gg x Gg. 2. One student in the group will be in charge of the male bag, the second student will be in charge of the female bag, and the third student will be the data keeper. 3. At the same time, each of the students controlling the bags of gametes, will reach into their bag and pull out one of the beads. The only possibilities that can be made from this selection are: GG, Gg, or gg. GG is homozygous green, Gg is heterozygous green, and gg is homozygous yellow. The third student will mark the resulting combination in the data sheet at the bottom of the page. 4. Return the beads back into the bag and conduct the same process 29 more times. Data Table

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Trial

Offspring's Genotype

Offspring's Phenotype

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 25 27 28 29 30 Summary: 1. What is the dominant trait? ______________.

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2. How do we know it is dominant? ____________________________________________________ 3. Which one is the recessive trait? ________________. 4. What is the genotype of the parents? __________________________________________________ 5. What is the phenotype of the parents? _________________________________________________ 6. Fill in the Punnett square below using the parents given in the procedure. lllllllllMale _________Gg________ qqqqqqqqX rrrrrrrrFemale _______Gg________

7. What is the genotypic ratio? ____________________. 8. What is the phenotypic ratio? ____________________.

©2000 Troy High School Biology

Corn Dihybrid Genetics Investigation*

Name___________________ 5

In this exercise you will study dihybrid inheritance by analyzing phenotypes of an F2 generation of corn grains. Obtain one ear of corn for each set of two members on your team. Do not remove any grains from the ears. Notice that some of the grains are purple and others are yellow. The purple color is produced by a pigmented layer within the grains. If the layer is not pigmented (is colorless), the yellow color of an inner tissue shows through. Sweet corn grains can be recognized because they wrinkle upon drying while starchy grains remain smooth. An individual grain may be purple and starchy, purple and sweet, yellow and starchy, or yellow and sweet. Recording Data Working in pairs, count and record in Table 1 the number of grains of each phenotype. One person should call out the phenotypes while the other records them in the table. Put a colored marker pin in the end of one row of grains and count and record the phenotypes. Put an uncolored row marker pin in the end of the next row and continue counting. After each row is completed, move the row marker pin to the next row until you return to the row marked by the colored pin. Exchange your data with another group and record (ear 2). When finished, combine the totals for each phenotype counted by your team and record them under “phenotype class” in Table 2. Then record the total number of grains counted. Data Interpretation Examine the totals obtained by your team. Which are the dominant phenotypes? ____________ and ______________. How do you know? The corn grains are the F2 generation resulting from a cross between a homozygous purple and starchy corn (R/R Su/Su) and a corn that is homozygous yellow and sweet (r/r su/su). To better understand how these ears were produced, fill in the following by placing the symbol for an allele in each blank. R = gene for color purple r = gene for colorless (yellow) Su = gene for starchy su = gene for sweet ____/____ ____/____ x ____/____ ____/____ Parents ____/____ ____/____ x ____/____ ____/____ F1 cross F1 x F1

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F2’s Thus we can see that the F1 gametes can combine in ______ ways to give rise to the F2, and that the expected phenotypic ratio of the F2 is: (fill in the phenotypes) 9 ________________ : 3 ________________ : 3 _________________ : 1____________ Using this information, compare your “number of individuals” (actual counts) with the number that would be expected for your sample size. To do this, divide the “Total Corn Grains Counted” by the number of possible gamete combinations (16) in the F2 generation. Multiply this number (the dividend) by, respectively, 9, 3, 3, and 1 to fill in the “expected numbers” of Table 2. Do the numbers of phenotypes you observed seem to be close approximations of the “expected numbers”? _________ What factors could cause variation from the “expected numbers”? What would be the genotypes and phenotype ratios expected in the following cross? R/r Su/su x r/r su/su ? Show your work.

Table 1 7

Purple Starchy

Purple Sweet

Yellow Starchy

Yellow Sweet

Total Ear 1 Total Ear 2 Total Ears 1 & 2

Table 2

Phenotype Class Number of Individuals (actual) Expected Number

1 Purple Starchy

2 Purple Sweet

3 Yellow Starchy

4 Yellow Sweet

a1

a2

a3

a4

e1

e2

e3

e4

4 Classes Total (1-4)

*Source: Carolina Biological Supply Company, “Corn Dihybrid Biokit”

Investigation: What is Your Genetic Profile? 8

Objectives: In this investigation you will (1) observe a variety of human traits; (2) collect and record data; (3) infer possible genotypes from your data and (4) calculate percentages. Materials: mirror Prelab Preparations: Review what you have learned about genetics by answering the following questions. (1) Define the terms phenotype, genotype, dominant, and recessive. (2) What is a pedigree? Why are pedigrees often necessary when studying human hereditary traits? (3) When reading a pedigree, how do you distinguish males from females and children from parents? Procedure: Making Your Genetic Profile 1. Form a cooperative group of two students. 2. Make a table (each student) similar to the first two columns in figure 1. 3. Working with your partner, check for each trait described below. Record your phenotype in the appropriate column in your table. a. Mid-digital hair Each segment of a finger is called a digit. Is hair present on the middle digit of any of your fingers? b. Tongue rolling Look in the mirror. When you stick out your tongue, can you roll up the edges on each side? c. Dimples When you smile, are there small indentions in your cheeks? d. Cleft chin Do you have an indentation in the middle of your chin? e.

Attached earlobes Do the tips of your ear lobes hang partially free, or are they completely attached to the side of your head? f.

Freckles Do you have small reddish- brown spots on your skin?

g.

Widow’s peak Pull your hair back over your forehead. Does the hairline come down to a short point in the middle of your forehead or does it go straight across?

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h.

Five fingers Were you born with five fingers on each hand?

Figure 1. ---------------------------------------------------------------------------------------------------------Trait | Phenotype | % of Class | Dominant | Genotype | | | | or Recessive | | ---------------------------------------------------------------------------------------------------------Mid-digital hair | | | | | ---------------------------------------------------------------------------------------------------------Tongue rolling | | | | | ---------------------------------------------------------------------------------------------------------Dimples | | | | | ---------------------------------------------------------------------------------------------------------Cleft chin | | | | | ---------------------------------------------------------------------------------------------------------Attached earlobes | | | | | ---------------------------------------------------------------------------------------------------------Freckles | | | | | ---------------------------------------------------------------------------------------------------------Widow’s peak | | | | | ---------------------------------------------------------------------------------------------------------Five fingers | | | | | ---------------------------------------------------------------------------------------------------------4. Compare your data with that of your class mates. Calculate the percentage of the class that exhibits each trait. 5. Look at the pedigree below. Colored figures represent individuals that possess the trait. Explain why individual IV-8 and her parents provide the evidence that attached earlobes is a recessive trait.

I

Attached Earlobes

II

III

IV

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6. Analyze the pedigrees for tongue rolling and five fingers. Explain whether each trait is dominant or recessive. Each of the other traits listed in your table is dominant.

I

Tongue Rolling

II

III

I

Five Fingers*

II 5

6

III

*In this case, the normal state of five fingers is not colored in. 6.1.Based on available information, use appropriate symbols for each trait to record your possible genotypes in your table. Analysis 1. Analyzing data Does the information you collected and studied during this investigation indicate that dominant traits are most common? Explain. 2. Evaluating Methods Look at the pedigree for five fingers. Explain why individuals II-5, II-6, and their children are the most important for analyzing whether this trait is dominant or recessive. Source (Holt Biology, Visualizing Life, George B. Johnson, 1994, Holt, Rinehart and Winston) 11

Id. #____ Name: ______________________________________ DG: 10/ DD: 10/ Turned in: ________ Lab # ____ Pedigree Study Grade: ___ / ____ = _____ % Introduction: Pedigrees are not reserved for show dogs and racing horses. All living things, including yourself, have pedigrees. A pedigree is a diagram that shows the occurrence and appearance, or phenotype, of a particular genetic trait from on e generation to the next in a family. Genotypes for individuals in a pedigree can usually be determined with knowledge of inheritance and probability. Over the past ten years nearly 200 elephants have died annually as a result of conflict with humans in Sri Lanka India. In India man and elephant have co-existed in the island from prehistoric times, and over the past 3,000 years the elephant has been considered an important and venerated cultural, religious icon. Experts believe that at the turn of the 19th century there were 20,000 elephants in Sri Lanka. The present population is less than 3,500 elephants. In 1998 over 350 elephants were killed in Sri Lanka. (www.bentghic.com/sri_lanka/issues.htm) In a woeful version of natural selection, or arguably artificial selection, ivory poaching may be causing Asian elephants to lose the gene that allows them to develop tusks. About 40 to 50 percent of the animals are normally tuskless, but in Sri Lanka, more than 90 percent of the population is not growing tusk. “When you have ivory poaching, the gene that selects for whether an elephant has tusks or not will be removed from the population,” said Paul Toyne, a species conservation officer on the WWF. (www.save-the-elephants.org) Objectives: Learn the meaning of all symbols and lines that are used in representing a pedigree. Predict geneotypes for all individuals shown in two sample pedigrees. Investigate real life cause, effect and solutions to environmental issues. Procedure: The pedigree in Figure 1 shows the pattern Of inheritance in a family of a specific trait. the trait being shown is elephant tusk. Geneticists Recognize two general tusk characteristics in Asian elephants. The males have visible tusk and the Females do not have visible tusk. The gene Responsible for Tusks (E) is dominate over The gene for no tusk (e). In a pedigree, each generation is represented by a Roman numeral. Each elephant in a generation is numbered. Thus each elephant can be identified by his/her generation numeral and number. Males are represented by square symbols while females are represented by round symbols (Figure 1) All darkened symbols on a pedigree are individuals who are homozygous recessive (ee) for the trait being studied. Therefore, elephants I-1 and II-2 have (ee) genotypes. They are the only two individuals who are homozygous recessive and show the recessive trait. They have no tusk.

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All undarkened symbols have at least one dominant gene (E). The genotype for person I-2 is either (EE) or (Ee). Punnett squares can be used to aid in determining the genotypes. If the mother (I-2) is EE, then a Punnett square such as the one in Figure 2 results in all children having tusk. Even though the Asian female does not have tusk she carries the genes for tusk in her DNA. 1) Do the pedigree symbols for the children support this? _____ (yes or no) 2) What are the Genotypes for children II-3 in figure 1? ________________________ 3) What are the genotypes for children II-4 in figure 1? ________________________ Since they have tusk, II-3 and II-4 can only be (Ee) because the father is “tuskless (ee)” and the mother has the genes for “tusk (Ee).” Because he has elephant II-1 (Husband of oldest daughter) may be (EE) or (Ee). All children be (Ee) tusked if the father were (EE) (Figure 2)

tusks, would

4) Does the pedigree support this? _____ (Yes or No) If the father were (Ee), half the children would probably have no tusk (ee) 3.

Figure

5) Does the pedigree support this? ____ (yes or no) There are not enough children to make a definite conclusion. With only one either genotype is possible for the father. Therefore, his genotype is shown as

child, (E).

Part II Using the same genes as above for elephant tusk inheritance, determine the genotypes of all individuals in family A (figure 4) and Family B (Figure 5). Record the genotypes of each elephant in Data table 1 be sure to include genotype (letters) and phenotype (physical characteristics, tusk or no tusk). IndividualGenotype Heterozygous Dominant Phenotype EE,Ee,ee Homozygous recessive“Tusker” Homozygous Dominant “no tusk” I-1 I-2 II-1 II-2 III-1 III-2 III-3 III-4 IV-1 IV-2

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The following area is to be used for Punnett squares.

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Family B (Figure B) The area below is to be used for Punnett squares: 29) 30) 31) 32) 33) 34) 35) 36) 37) 38) 39) 40) 41) 42) 43) 44) 45) 46) 47)

IndividualGenotype Heterozygous Dominant Phenotype EE,Ee,ee Homozygous recessive“Tusker” Homozygous Dominant “no tusk” I-1 I-2 II-1 II-2 II-3 II-4 III-1 III-2 III-3 III-4 III-5 III-6 III-7 III-8 III-9 III-10 III-11 III-12 IV-1

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Analysis: Investigate the following recopied figure of Family A. Answer the questions using your knowledge of genetics and facts from the lab to support your answers. Use the area below for Punnett square work.

6) What are the three changes to this copy of Figure 4 Family A? ___________ _________________________________ _________________________________ _________________________________ _________________________________ Fill in the data table for each individual of this new family of Asian elephants

52) 53) 54) 55) 56) 57) 58) 59) 60) 61)

Individual Genotype Heterozygous Dominant Phenotype EE,Ee,ee Homozygous recessive“Tusker” Homozygous Dominant “no tusk” I-1 I-2 II-1 II-2 III-1 III-2 III-3 III-4 IV-1 IV-2

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Part III The following is the family tree for the Rosamond Gifford zoo Asian elephant breeding program.

Elephant

Sex

Parents

Year Birth

Living location

Indy

Male

?

Wild caught

Rosamond Gifford Syracuse NY

Targa Romani Mali Stillborn Tundi

Female Female Female Female Male

? ? Indy/Targa Indy/Targa Indy/Romani

Wild caught Wild caught 1997 1999 1991

Rosamond Gifford

Kirina Preya Shanti Kumari Kundulah

Female Female Female Female Male

Indy/Romani Indy/Romani ? Indy/Shanti Shanti/?

2000 ? 2000 2002

zoo

zoo Rosamond Gifford zoo Rosamond Gifford zoo Died at birth Wipshade zoo Europe Rosamond Gifford zoo Died of Herpes/2003 Buffalo zoo Died of Herpes Lived

65 – 80) Create a Pedigree chart for Indy’s family. Be sure to use the rules of a pedigree chart to indicate the following: 1) Sex of the individual 2) Generation of the individual 3) connect family relationships You do not have to do the Genotypes or Phenotypes of the Elephants in this family tree.

Application questions: 7) How many offspring does Indy have? _________________________________________________________

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8) Who are the daughters of Romani? ___________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ 9) Would it be a good idea to mate Karina with Tundi? _______ 10) Explain why this would not be a good idea. __________________________________________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ __________________________________________________________________________________________ 11) Asian elephants in the wild are considered an endangered species, what does this mean?

_____________________________________________________________________________ _____________________________________________________________________________ _____________________________________________________________________________ The Rosamond Gifford zoo elephant family helps educate people about elephant endangerment everyday. They also breed the elephants to increase the population that is in captivity around the world. Explain two reasons you think it is important to be breeding elephants in captivity. 12) Reason # 1

_____________________________________________________________________________ _____________________________________________________________________________ _____________________________________________________________________________ _____________________________________________________________________________ _____________________________________________________________________________ 13) Reason # 2 __________________________________________________________________________________________

_____________________________________________________________________________ _____________________________________________________________________________ _____________________________________________________________________________ _____________________________________________________________________________ 14) The Rosamond Gifford zoo participates in a Species Survival program. In this program a Pedigree Chart of all the elephants in captivity is kept. Explain using facts and things you have learned from this lab how this could someday benefit the plight of the Asian elephants in Sri Lanka India. __________________________________________________________________________________________

_____________________________________________________________________________ _____________________________________________________________________________ _____________________________________________________________________________ _____________________________________________________________________________ _____________________________________________________________________________ _____________________________________________________________________________ _____________________________________________________________________________ 18

_____________________________________________________________________________ _____________________________________________________________________________ _____________________________________________________________________________ PURPOSE  To learn how variability in facial characteristics is achieved through a random assortment of  dominant and recessive alleles.

INTRODUCTION Human variation in appearance is due not only because of the large variety of traits that exist  in a population but also because of the random mixing of alleles that occurs during sexual  reproduction. Each parent contributes half of each child’s genetic make­up. In other words, each  parent contributes one allele for each gene locus. Several different patterns of inheritance are possible. Many alleles are dominant, which  means they mask those that are recessive. Recessive alleles can be expressed only when  dominant alleles are not present. In some cases, incomplete dominance prevents the  expression of either allele, resulting in an intermediate phenotype. Other traits, such as beard  growth, are gender specific, meaning that the expression of such a trait is limited to one of the two  sexes. Other human traits, such as height or skin color, show continuous variation in the population.  These traits are thought to be controlled by many genes and are called polygenic traits. Some  traits are controlled by interaction between multiple genes. One such interaction is called  epistasis. Epistasis occurs when genes at one location affect the expression of genes at another  location. For example, an individual may have two dominant genes for the production of melanin  (a pigment that gives skin color). If the individual also has two recessive alleles for albinism, the  genes for melanin will be silenced. The activity below demonstrates how each of these modes of inheritance influence  phenotype. “Parents” will flip a coin to simulate the role of probability in the independent  assortment of chromosomes during gamete formation. The two parents will work together to  determine the facial characteristics of their offspring. Once the coin­flipping activity has been  completed, you will draw a picture of what your child would look like in his or her teens. Your  child’s picture should be sketched on an overhead transparency. At the end of lab, each set of  parents will present their child to the class and explain how the genotype of their child results in  the phenotype shown.

MATERIALS  two coins  transparency sheet  transparency marker

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Source: http://homepage.mac.com/massasoit.bio/faculty/lab12a/lab12a.html PROCEDURE 1.

Pair up with a classmate. You will each act as a “parent” to this fictitious “child.” Begin the  simulation with the assumption that each of you has one dominant and one recessive  allele (i.e., you are both heterozygotes) for each of the facial features illustrated on the  following pages.

2. To determine the genotype inherited by your child, both you and your partner will flip your  coins. Heads represents dominant alleles; tails represents recessive alleles.

Example: If your partner flips heads and you flip tails, your child’s genotype for that trait  would be heterozygous (one dominant and one recessive allele).

3. First, determine the sex of the child. Which parent should flip the coin? In humans, the  fertilizing sperm (from the male) determines the sex of the child. All eggs have X  chromosomes, but roughly half the sperm produced by the father will have an X  chromosome and half have a Y chromosome. If a Y­bearing sperm fertilizes the egg, the  child will be male (XY). If an X­bearing sperm fertilizes the egg, the child will be female  (XX). Therefore, in this simulation, the father will flip a coin to determine the sex of the  child. If the father flips a head, your child is a boy; if tails are flipped, your child is a girl.

4. Give your child a name. Record this information on your lab report.

5. You and your partner should simultaneously flip coins for each facial feature. Record the  genetic contribution of the parents and the offspring’s genotype and phenotype on the lab  report.

6. Draw your child on the transparency provided and be prepared to describe him/her to the  class. 7. 8. After you have completed steps 1 – 6, summarize below, what you have learned  concerning human inheritance.

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Parents’ names:

_____________________ and ________________________

Child’s sex: ________ _____________________

Trait

Child’s name:

Allele 

Allele 

from  Dad

from  Mom

Child’s  genotype

Child’s phenotype

1. Face shape 2. Chin size 3. Chin shape 4. Cleft chin

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5. Skin color 6. Hair type 7. Widow’s peak 8. Eyebrow color 9. Eyebrow thickness 10. Eyebrow placement 11. Eye color 12. Eye distance 13. Eye size 14. Eye shape 15. Eye slantedness 16. Eyelashes 17. Mouth size 18. Lips 19. Protruding lip 20. Dimples 21. Nose size 22. Nose shape 23. Nostril shape 24. Earlobe attachment 25. Darwin’s earpoint 26. Ear pits 27. Hairy ears 28. Freckles on cheeks 29. Freckles on forehead

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23

24

25

26

27

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Strawberry DNA Extraction Purpose: To experience the unbridled joy of actually seeing DNA. Seriously though, extraction of DNA is required for study and manipulation of DNA, including genetic engineering, DNA fingerprinting (such as for crime investigation), genetic testing, gene sequencing, genome sequencing, and many more applications. Strawberries are octoploid, so they are excellent sources of DNA. Solutions: DNA extraction buffer: per L of water: 146 g NaCl, 50 mL dishwashing detergent (or shampoo), 95% ethanol, cold Materials: Ziploc bag, one 3mL pipette, two cuttings of cheesecloth, one stick, and one new 15 mL tube per pair. You should have at your table another 15 mL tube (for DNA extraction buffer; funnel and 50 mL graduated cylinder Overview: A strawberry will be mashed to break up cells walls. Then DNA extraction buffer (essentially soapy saltwater) will be added. The soap will break up cell membranes, and the salt will help get much of the cell debris (including many of the proteins) to precipitate out of the solution. The DNA will stay mostly in solution in the extraction buffer. Filtering will remove cell debris. Cold 95% ethanol will then be added over some of the filtered strawberry DNA extract. Since DNA is mostly insoluble in ethanol it will precipitate. Although any one DNA strand is too thin to see with the naked eye, several strands will clump together and be visible. Procedure: 1. Work in pairs. Obtain one strawberry, one Ziploc bag, one 3mL pipette, two cuttings of cheesecloth, one stick, and one new 15 mL tube per pair. You should have at your table another 15 mL tube (for DNA extraction buffer; this one is reused each lab), a rack for tubes, one standing 50 mL tube, and one funnel. There should also be a squeeze bottle of DNA extraction buffer at your table (shared by the whole table). 2. Put 10 mL of DNA extraction buffer into the appropriate tube at your table (it should be labeled “DNA Ex. Buffer”). Use the 10 mL line to tell when you have enough. Don’t count the soap bubbles as part of the buffer, and don’t worry about being perfect – close to 10 mL will do. 3. Take the green sepals off the top of the strawberry and put the strawberry in the Ziploc bag. Close the bag, squeezing out most of the air as you do so. Just throw the sepals away. 4. Being careful not to break or open the back, thoroughly mash the strawberry for two minutes.

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5. Pour the 10 mL of DNA extraction buffer into the back, reseal the bag, and mash the strawberry for 1 minute, mixing it with the DNA extraction buffer. Put the 15 mL “DNA Ex. Buffer” tube back in the rack for the next class; do not clean or rinse this tube. 6. Place the funnel in the standing 50 mL tube, and place the cheesecloth in the funnel so that it makes two layers. Push the cheesecloth in a bit with your fingers, but don’t worry about getting all of the way in – the strawberry material will take care of that for you when you pour it on the cheesecloth in the next step. 7. While one person holds the tube, the other should carefully pour the mashed strawberry DNA extract onto the cheesecloth in the funnel. The DNA extract will drip through. 8. After about 2 minutes the dripping should have slowed; you will likely have 10 mL or more of filtered strawberry DNA extract in the 50 mL tube. All you need is 2 mL. 9. Throw away the cheesecloth, set aside the funnel, and carefully pour 2 mL of the filtered strawberry DNA extract into the clean 15 mL tube (use the lines on the tube to measure). Try to get close to 2 mL, but a bit more is fine. Put the funnel back in the 50 mL tube to help keep things clean. 10. By this time the TAs should have provided each group with a small bottle of cold 95% ethanol. Open the ethanol and use the pipette to get 3 mL of ethanol. If you don’t have 3 mL try again – a bit less or more is fine, but try to be pretty close. 11. Hold the 15 mL tube with the strawberry DNA extract at an angle and slowly add the 3 mL of ethanol down the side of the tube. Be careful not to disrupt the layer that forms between the strawberry DNA extract and the ethanol. 12. Watch closely as the DNA precipitates and clumps together at the layer between the extract and the ethanol. Tiny bubbles will appear as the DNA precipitates. 13. After 1-2 mins of precipitation you should have plenty of DNA. Put the stick into the ethanol layer and slowly and carefully rotate it to wind or “spool” the DNA onto the stick. Then carefully remove the stick and look at the DNA. Cleanup: The 50 mL standing tube and funnel should be rinsed out and placed back at your table for the next lab. Place the ethanol container where indicated by the TAs. The 15 mL tube that was used for DNA precipitation should be rinsed and placed in the cleaning bin indicated by the TAs. The stick, Ziploc bag, and 3 mL pipette should be thrown away. Follow up questions: 1. What cell barriers had to be disrupted to get the DNA out, and how were those barriers disrupted?

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2. What caused the DNA to precipitate? 3. What does the DNA look like to you? 4. Give at least one reason that scientists, doctors, or others would extract DNA as part of their job (not just that it is a “neat thing to do”).

http://www.auburn.edu/academic/classes/biol/1021/lab12DNA.htm

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