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Each parent contributeshalf of the offspring's genes."Special cells, called GAMETES, arenecessary for this to happen.Gametes -- eggs in females and"sperm in males -- contain half the"chromosomes of regular cells. They"are created through a special type of cell division called MEIOSIS."In humans, regular body cells have 46 chromosomes. How many do eggsand sperm have?Type number and press RETURN."In fruit flies, regular body cellshave 8 chromosomes. How many doeggs and sperm have?462384!No matter how many chromosomes an#organism has, its gametes have half'that number. At fertilization, a gamete#from one parent joins with a gamete&from the other. Together, they restorethe full number of chromosomes.Cells with the full number ofchromosomes are called DIPLOID. Gametes are HAPLOID. Let's watch"what happens to the chromosomes of!a fruit fly as the cell undergoesmeiosis.ORGANISM BODY CELLS ChimpanzeeCucumber Hermit crabMosquitoPotato Snapdragon4814254648162471273248GAMETESDIPLOIDHAPLOID$Meiosis involves two cell divisions.'During INTERPHASE I, the cell's geneticmaterial doubles.disappears. Each chromosome and its replicate, called the SISTER$CHROMATID, attach at the CENTROMERE.#Next, each doubled chromosome finds another doubled chromosome. Thisis the PAIRING OF HOMOLOGOUS CHROMOSOMES. The HOMOLOGS form a#TETRAD. A tetrad is a group of four$chromatids joined at the centromere.#At PROPHASE I, the nuclear membrane%During Metaphase I, the tetrads alignalong the METAPHASE PLATE. Two"CENTRIOLES have moved to each pole!of the cell. In addition, SPINDLEFIBERS have formed. SKIP LINE In ANAPHASE I, the next phase of!meiosis I, the homologs separate."They move along the spindle fiberstoward the centrioles.%Four chromosomes go in one direction.Their homologs go in the other.$Finally, at TELOPHASE I, the chromo-#somes uncoil. A membrane forms sep- arating the cells. The cytoplasm!divides in two. Each new cell hasfour doubled chromosomes.These cells divide, but no newchromosomes are made. First,the chromosomes condense. Are they:DoubleRight.'The chromosomes are still double. Next,#during METAPHASE II, the centrioles%move to the poles and the chromosomes"line up along the metaphase plate.SinglePress RETURN to see."What forms between the centrioles? ChromatidsSpindle fibersTetrads"Good. The doubled chromosomes have"aligned along the metaphase plate. Next, the single chromatids moveSpindle fibers form between the#centrioles. The doubled chromosomes have aligned along the metaphaseplate."During Anaphase II, the chromatids#will move toward the poles. Nuclear$membranes will form, and the chromo-somes will decondense.#At TELOPHASE II, new cell membranesform. Press RETURN to see.How many chromosomes are now in each cell? EMPTY LINE"Right. These cells will become thegametes. Are the gametes:HaploidDiploid#Each cell now has four chromosomes.%These cells become gametes. Are they: fruit fly: That's right. Gametes are alwayshaploid. Gametes are always haploid. Theyhave half the regular number of chromosomes.#So far, we have concentrated on thedistribution of chromosomes in#meiosis. Let's consider the rest ofthe cell. What that looks like#depends on whether we are observingmeiosis in males or females.!SPERMATOGENESIS is the process of#sperm production in males. Watch asmeiosis proceeds.#Has the cytoplasm been distributed:Equally UnequallyRight. In spermatogenesis, thecytoplasm divides equally. The"result is four equal sperm. Notice that the sperm develop tails andchange shape as they mature.No. In spermatogenesis, the"Let's contrast meiosis in the malewith meiosis in the female."This process, called OOGENESIS, isdifferent in that the cytoplasmdoes not divide equally.$One cell gets most of the cytoplasm,and hence the organelles in it.!The other cell contains a nucleus"and only a bit of cytoplasm. It is!called a POLAR BODY. It will die. will die. What does the female gamete, thebig cell, become: A polar bodyAn egg#Sure. The big cell becomes the egg.$When fertilization occurs, the sperm"The sperm's nucleus enters the eggand the two nuclei fuse.How many chromosomes does thefertilized egg have?!No. The big cell becomes the egg.$Right. How many chromosomes does thefertilized human egg have?Since each gamete had four, the#fertilized fruit fly egg has eight.#How many chromosomes does the humanfertilized egg have?Right.The fertilized human egg has 46 chromosomes.triggers completion of meiosis.Press RETURN to see.$As this egg develops into an embryo,an infant, a child, and then an#adult, its genes will be expressed."Some characteristics will resemble the mother, some the father, andsome neither parent at all. Prophase I Metaphase I Anaphase I Telophase I Metaphase I Metaphase IIMaleFemale Polar bodyEggSperm Anaphase II Fruit fly Interphase I Telophase IIHere's a challenge. Look at thepeople pictured above. How many"traits can you think of that these people have inherited from their parents? Enter your number here,then press RETURN: Pretty good.Not bad.That's good. Each individualBut in reality, each individual inherits thousands and thousands"of traits from his or her parents.#You may have thought of some of the%traits identified above. Press RETURNto see some you may not have considered.All organisms, not just humans,"inherit traits from their parents.The study of heredity is called GENETICS.%Our earliest scientific insights into"heredity came from Gregor Mendel'sstudies of pea plants. SKIP LINE SKIP LINE SKIP LINELook at the pea plants pictured!above. Some are tall and some are SKIP LINE$short. Some have flowers just at the%tips; some have flowers all along thestem.#Now look at the pea seeds. Some are"wrinkled and some are smooth. Some$are yellow (white here) and some aregreen.Genes TemperatureWhat causes these differences? SKIP LINE. SKIP LINE#Right. All organisms that reproducesexually receive one gene for a&given trait from each parent. Meiosis,#a special process of cell division,makes this possible. SKIP LINE No. All organisms that reproduce$We call the forms of a gene ALLELES.%Tall and short are two alleles of the$gene that governs height. Seed color$is also governed by two alleles. Oneis yellow. What is the other?No. Green and yellow are twoRight. Green and yellow are two!alleles for the gene that governs seed color.#Let's see how genes are passed fromparents to offsprings. SKIP LINE Press RETURN to see one possible!result of crossing these two tall pea plants.%All the offspring are tall. Now press#RETURN to see what happens when youcross two short pea plants.&All the offspring are short. Now press#RETURN to see what happens when you!cross one tall pea plant with oneshort pea plant.#Do you think that all the offspringare tall because: Of chance."Tallness dominates over shortness."No. Some alleles for a given trait%Right. Some alleles for a given trait SKIP LINE%are DOMINANT. If a dominant allele is%present, it will always be expressed. The unexpressed allele is called RECESSIVE."Geneticists use letters as symbols'for alleles. Upper case letters are for'dominant alleles. Lower case designatesrecessive alleles. Is the aboveorganism tall or short? EMPTY LINE!Right. The organism is tall sinceNo. The organism is tall since!one gene for tall is present, andtall (T) is dominant.Geneticists also use symbols to$designate the various generations oforganisms under study."P@1 designates the parental gener-ation. F@1 designates the first"generation of offspring. It is the#first FILIAL generation. The second#filial generation is F@2, the third F@3, etc.%What do you think this F@2 generationwill look like? All tall. All short.Some tall and some short.!Right. Each of the F@2 plants got#No. Each of the F@2 plants receivedone allele for height from eachparent. Press RETURN to see thepossible genotypes.EMPTYEMPTYEMPTYEMPTYEMPTYEMPTYEMPTY Approximately one-quarter of the!offspring will be short. The rest will be tall."ance of an organism its PHENOTYPE.They call an organism's geneticmakeup its GENOTYPE."What will be the phenotype of this pea plant?TallShortRight.No.#Organisms that have two of the same#allele for a trait are called HOMO-#ZYGOUS. Organisms with one dominant$and one recessive allele for a traitare called HETEROZYGOUS.$Geneticists call the outward appear-As Mendel mapped patterns of!inheritance in his pea plants, he$faced a question: How could a parentpass on just one allele to anoffspring when it had two?"To answer this question, he formed"the above hypothesis. It is calledMendel's LAW OF SEGREGATION.LAW OF SEGREGATIONA pair of factors [genes] is$segregated, or separated, during theformation of gametes.What process of cell divisionconfirms Mendel's hypothesis?MeiosisMitosis$Right. Meiosis reduces the number ofgenes in a cell by one half.!No. MEIOSIS reduces the number ofHere's an interesting question:Is the inheritance of one trait$linked to the inheritance of anothertrait?$Look at the height and at the place-%ment of the flowers on the two plants#pictured above. Press RETURN to seetheir offspring.#Does the inheritance of height seem$related to the inheritance of flowerplacement (Y/N)?%Right. If inheritance of these traits"No. If inheritance of these traits"were related, any tall plant wouldhave terminal flowers.$Many, but not all, traits are inher-"ited independently of one another.%We now know that this is true as long"as the genes for the traits are ondifferent chromosomes.LAW OF INDEPENDENT ASSORTMENTFactors [genes] are inheritedindependently of one another.%Scientists today know much more about"organisms and inheritance than did$Gregor Mendel. But the ideas coveredhere are still called MENDELIAN"GENETICS after the man who was thefather of genetics. Hair color Hair textureBaldness Skin color Eye colorMyopiaEye size Shape of faceHeight Frame size Blood typeColor blindness Rh FactorSickle cell anemiaEnzymesSugar toleranceMilk tolerance Lung capacityHeart functionBlood clottingType answer, then press RETURN. Male gameteEggGenes ChromosomesTTttTt?P@1F@1F@2F@3 Heterozygous HomozygousDominant Recessive PhenotypeGenotype Flat feet Finger printsFrecklesTooth structure Bone length Torso heightInsulinGrowth hormoneBlood cholesterolVaricose veins Skin textureMetabolic rateAlcohol toleranceAlergies"You've probably seen them. They're$the tiny flies that buzz around ripe fruit. We call them fruit flies.Scientists call them Drosophilamelanogaster. They are an ideal#organism for the study of genetics.#Fruit flies are ideal organisms for!geneticists to study because they produce new generations quickly.#Which of the organisms listed above"do you think is the most difficult to study?OrganismGeneration timeCornE. coli Fruit flyHumansPink bread mold Pea plants6 months 20 minutes10 days20 years14 days6 months#That's right. The long time between$human generations plus the vast num-"ber of human genes (between 50,000$and 100,000) complicate the study ofhuman genetics."No. Humans are the most difficult.#The long time between human genera-#tions plus the vast number of human"genes (between 50,000 and 100,000)complicate the study of human genetics.%The short time between generations is%only one reason why geneticists studyfruit flies. In addition:Fruit flies are easy to breed.They grow fast.They produce many offspring. Let's take a closer look at this humble subject of genetic study.Which fruit fly is bigger? EMPTY LINE Right. The female has a slightlylarger body than the male.No. The female has a slightlyWhich fly has a larger abdomen?"Yes. The female also has a longer,more pointed abdomen. sex has a small tuft of bristleson its foreleg?!No. The female has a longer, morepointed abdomen."Now look at the flies' legs. Which%Look again. The male has a small tuft"of bristles, called a SEX COMB, on its forelegs. Right. The male has a small tuft%The male's abdomen is also darker and$has fewer stripes than the female's.$These differences enable geneticiststo distinguish easily males and#females -- an obvious necessity forexperimental matings!%Now you try to determine the sex of a!fruit fly. Press RETURN to selecta fly. SKIP LINE Is your fly:MaleFemale Right. You're on your way to the genetics lab!#No. Remember that the female fly is larger and has a longer, lighter$colored abdomen than the male. Also,#the female does not have a sex combon her forelegs.$No. Remember that the male fly has a$shorter, darker colored abdomen than#the female. The male also has a sexcomb on his forelegs.&Eye color and wing size are but two of#the traits geneticists can study in'fruit flies. Others? Hairy wings, eagle'(extended) wings, small eyes, bar eyes,&no eyes, and no wings. And that's justeyes and wings!#the egg. It is white and segmented.%It is eyeless and legless. But it hasa mouth. It eats. And it grows.#DAY 3: The larva sheds its skin and%mouthparts. It enters a second larval stage. A third larval stage will follow in another day. The larvacontinues to eat and grow.$DAY 6: The third larval skin darkens%and hardens. It becomes a cocoon-like$PUPA. The pupa protects the larva as it transforms into an adult fly.!DAY 10: The adult fly emerges. It will live from 40 to 70 days. If$female, it will start laying eggs in!two days. During its lifetime, itwill lay almost 3,000 eggs. These large generations of fruit"flies provide ample populations in%which to study inheritance. Eye color$is one trait geneticists study. Most%fruit flies have bright red eyes. Butsome have black eyes.Wing size is another trait. Two!alleles govern wing size in fruit$flies. Most wings are long. But some flies have "dumpy," short wings.Understanding how traits arepassed on in a simple organism like the fruit fly is important.!It gives geneticists insight intoinheritance in other species,including your own.Use @R @L to move cursor, entergamete, and press RETURN.$Use @U @D @R @L to move cursor, typegamete, then press RETURN. Enter genotype and press RETURN.#Enter number from 0 to 16 and pressRETURN.!This is a Punnett square. It is a"tool geneticists use to figure outthe result of crosses.$Geneticists plot all of the possible#gametes of the female parent acrossthe top of the square. Press @R#to see the gametes for a homozygoustall (TT) female pea plant.%They plot all of the possible gametes$of the male parent along the side ofthe square. Press @R to see the$gametes for a heterozygous tall male pea plant.$To fill in the Punnett square, gene-%ticists carry the male gametes across"all of the boxes. Press @R to see. #They carry the female gametes down.Press @R to see.$Note that the capital letters always%appear first. This is a convention ofgenetic notation.%The completed Punnett square provides#important information. It shows all!possible offspring. It also shows$the probability of each type of off-spring occuring.#In the above example, 50 percent of$the offspring will be homozygous for"tall. The other 50 percent will beheterozygous for tall.%How many of the offspring will appeartall? 100 percent 50 percent#Right. All of the offspring will be"tall. You can see that geneticists"can use Punnett squares to predict phenotypes as well as genotypes. No. All of the offspring will be%Would you like to practice filling ina Punnett square (Y/N)?!Fill in the gametes. Press D whendone.#No. The female gametes go along the top. They are T and t. The male$gametes go down the side. They are tand t.#No. The female gametes go along thetop. They are T and t. No. The male gametes go down theside. They are t and t.%Do you want to try filling in anotherset of gametes (Y/N)?#Now fill in each box of the Punnettsquare. Press D when done."The highlighted squares are incor- rect. Remember to bring the male#gametes across all the boxes. Bring!the female gametes down. Press @Rto see.!Do you want to try filling in theoffspring again (Y/N)?!How many of the squares representtall offspring?Right.No. Two squares represent tall offspring.%No. All of the squares represent tall offspring.$The ratio of tall to short offspringfor this cross is 1:1."All of the offspring will be tall.#No. The female gametes go along thetop. They are t and t.!No. The male gametes go along theside. They are T and T.#No. The female gametes go along the'top. They are t and t. The male gametes$go along the side. They are T and T.#You've just used the Punnett square to predict offspring showing one%trait. You can use the same technique$for two, three, or even more traits.$Each time you add another trait, the%number of possible gametes increases.The Punnett square becomes more"complex, but the principles remain#the same. Let's examine the exampleabove.#These parents are both heterozygous"tall pea plants with axial flowers(TtAa).#The first female gamete is TA. Fill!in the rest. Press D when you aredone.No. You must join each possibleallele for one trait with each%possible allele for the second trait.Press @R to see.#Now try the male gametes. The first$one is TA. Fill in the rest. Press Dwhen you are done.#Good. Now fill in the male gametes.Press D when you are done.%Fill in the genotype for this square.Right. How about this square?%No. Remember, you must bring the male#gamete across and the female gamete#down to each block. Press @R to seethe correct genotype.Now try this one.$Here's the completed Punnett square.%Count the number of boxes that repre-sent tall offspring with axialflowers.%Good. Now count the boxes that repre-!sent tall offspring with terminalflowers.%No. Tall offspring with axial flowers can have either of the following#genotypes: TTAA or TtAa. Nine boxescontain these genotypes."Now count the boxes that represent%tall offspring with terminal flowers.%Right. How many boxes represent shortoffspring with axial flowers?#Right. And how many boxes representshort offspring with terminalflowers? Very good. No. Tall offspring with terminalflowers can have either of the"following genotypes: TTaa or Ttaa.$Three boxes contain these genotypes.No. Short offspring with axialflowers can have either of the"following genotypes: ttAA of ttAa.$Three boxes contain these genotypes.!No. Short offspring with terminal%flowers have the genotypes ttaa. Onlyone box contains this genotype.How many boxes represent shortoffspring with axial flowers?How many boxes represent short offspring with terminal flowers?"By counting the boxes representing$each possible type of offspring, you%have figured out the PHENOTYPIC RATIOfor this cross. It is 9:3:3:1.TRAITSMATESPUNNETTCROSSEXIT?!Press @U @D to see choices. PressRETURN to select.Press @R to continue.$Use @U @D @L @R to move cursor, typeletter(s), then press RETURN.Press D when done."The highlighted entries are wrong.Do you need help (Y/N)?LAB: PUNNETT SQUARES You can predict the genotype and"phenotype of offspring if you know%the parents' genotypes. This lab lets"you use a special tool, called the%PUNNETT SQUARE, to make these predic-%tions. You will work with pea plants."Do you want instructions on how touse the Punnett square (Y/N)?$Select the number of traits that youwant to use in your crosses: One trait Two traits Three traits$PURPOSE: To predict the frequency of&a trait in the offspring of pea plantsof known genotype&RATIONALE: The Punnett square lets you#determine the theoretical number of%offspring having different genotypes.$Once you know the genotypes, you can%determine the offsprings' phenotypes.!You can compare these theoreticalvalues to actual offspring. PROCEDURE:1. Select trait.2. Select genotypes of mates.3. Fill in Punnett square.4. Select the phenotypic ratio.6. Cross the peas.%7. Count offspring of each phenotype.%8. Compare the actual results to your hypothesis.RATIONALE: You will follow thisprocess:#1. Predict: The Punnett square lets&you predict the number of offspring ofdifferent genotypes.$2. Hypothesize: After predicting the&outcome, you will choose a hypothesis."3. Experiment: The cross shows the#results of actual field experiments'against which you test your hypothesis. 5. Select a hypothesis about theoutcome of an actual cross.$PURPOSE: To predict the frequency of"two traits in the offspring of peaplants of known genotype$Purpose: To predict the frequency of three traits in the offspring ofparents of known genotype1. Select traits.HEIGHTTall (T) is dominant.Short (t) is recessive.FLOWER POSITIONAxial (A) is dominant.Terminal (a) is recessive. SEED COLORYellow (Y) is dominant.Green (y) is recessive. SEED FORMRound (R) is dominant.Wrinkled (r) is recessive.Tall: TTTall: Tt Short: tt Axial: AA Axial: Aa Terminal: aa Yellow: YY Yellow: Yy Green: yy Round: RR Round: Rr Wrinkled: rrTall/axial: TTAATall/axial: TtAATall/axial: TTAaTall/axial: TtAaTall/terminal: TTaaTall/terminal: TtaaShort/axial: ttAAShort/axial: ttAaShort/terminal: ttaaSelect a female parent.Now select a male parent.Tall/yellow: TTYyTall/yellow: TtYyTall/green: TTyyTall/green: TtyyShort/yellow: ttYYShort/yellow: ttYyShort/green: ttyyAxial/yellow: AAYYAxial/yellow: AaYYAxial/yellow: AAYyAxial/yellow: AaYyAxial/green: AAyyAxial/green: AayyTerminal/yellow: aaYYTerminal/yellow: aaYyTerminal/green: aayyTall/round: TTRRTall/round: TtRRTall/round: TTRrTall/round: TtRrTall/wrinkled: TTrrTall/wrinkled: TtrrShort/round: ttRRShort/round: ttRrShort/wrinkled: ttrrAxial/round: AARRAxial/round: AaRRAxial/round: AARrAxial/round: AaRrAxial/wrinkled: AArrAxial/wrinkled: AarrTerminal/round: aaRRTerminal/round: aaRrTerminal/wrinkled: aarrYellow/round: YYRRYellow/round: YyRRYellow/round: YYRrYellow/round: YyRrYellow/wrinkled: YYrrYellow/wrinkled: YyrrGreen/round: yyRRGreen/round: yyRrGreen/wrinkled: yyrrTTTtttAAAaaaYYYyyyRRRrrrTTAATTAaTtAATtAaTTaaTtaattAAttAattaaTTYYTtYYTTYyTtYyTTyyTtyyttYYttYyttyyAAYYAaYYAAYyAaYyAAyyAayyaaYYaaYyaayyTTRRTtRRTTRrTtRrTTrrTtrrttRRttRrttrrAARRAaRRAARrAaRrAArrAarraaRRaaRraarrYYRRYyRRYYRrYyRrYYrrYyrryyRRyyRryyrr$You must select both a trait and twomates before you can run the experiment.!Fill in the gametes. Press D when you are done.%Gametes for this trait are designated"by the two letters T and t. Pleasereenter."The genotype for the (male/female)you selected is (TT/Tt/tt).Please reenter the gametes.$You have not entered the gametes. Do you want to:Fill in the Punnett squareEnter the phenotypic ratioSelect a hypothesis to testThose still are not the right%gametes. Would you like to review the instructions for using a Punnett square (Y/N)?Fill in each box of the Punnett"square. Press D when you are done.Genotypes for this trait are"designated by a combination of the letters T and t, e.g., Tt or TT.Please reenter.$The highlighted boxes are not right."Do you want to correct them (Y/N)?"Reenter genotype and press RETURN. PhenotypeNumberTallShortAxialTerminalYellowGreenRoundWrinkled Tall/axial Tall/terminal Short/axialShort/terminal Tall/yellow Tall/green Short/yellow Short/green Axial/yellow Axial/greenTerminal/yellowTerminal/green Tall/round Tall/wrinkled Short/roundShort/wrinkled Axial/roundAxial/wrinkledTerminal/roundTerminal/wrinkled Yellow/roundYellow/wrinkled Green/roundGreen/wrinkled Select the phenotypic ratio thatreflects these results:1. 1:02. 0:13. 1:14. 3:1 This ratio suggests which of thefollowing hypotheses:$All offspring will show the dominanttrait.%All offspring will show the recessivetrait.Half of the offspring will show%the dominant trait and half will showthe recessive trait.$Three-quarters of the offspring will show the dominant trait and one-&quarter will show the recessive trait.trait.!One-quarter of the offspring willshow the dominant trait andthree-quarters will show therecessive trait.$9/16 of the offspring will show bothdominant traits.$9/16 of the offspring will show bothrecessive traits.#3/16 of the offspring will show one!dominant and one recessive trait.$1/16 of the offspring will show bothrecessive traits."Do you want to do an experiment totest this hypothesis (Y/N)?Count the offspring of each phenotype.#Are there differences between males$and females? At the chromosome levelthere sure are.$Look at this representation of human!chromosomes. Twenty-two pairs are"matched in both males and females.These are called AUTOSOMES.%The last pair, however, is matched in%females but not in males. These chro-&mosomes determine sex. They are called#SEX CHROMOSOMES. Females have two X!chromosomes. Males have one X andone Y.%Do you think that the sex chromosomes!carry genes for any traits except gender (Y/N)? No. The sex chromosomes do carry#Right. The sex chromosomes do carry&genes for other traits. However, the Y#chromosome carries fewer genes than"the X chromosome. Therefore, males"have only one gene for some traitswhile females have two.%Color blindness is an example of such&a SEX-LINKED TRAIT. It is carried only%on X chromosomes. Let's see how it is inherited.%We can represent normal vision with a!C and color blindness with a c. Awoman with normal vision is%X@WC@ZX@WC@Z. Fill in her gametes forthe Punnett square.Type gamete and press RETURN. No. The gametes are both X@WC@Z.#Now fill in the gametes for a color blind man.!Good. Fill in the Punnett square.!No. Remember that a man has one X%chromosome and one Y chromosome. Only#the X chromosome carries a gene for"color blindness. Since this man is'color blind, his gametes are X@Wc@Z andY.Now fill in the Punnett square.Right.$Try again. The highlighted boxes arewrong. No. The correct genotypes appearabove. No. The correct genotype appearsabove.GENOTYPE PHENOTYPE X@WC@ZX@Wc@ZX@WC@ZYX@Wc@ZYWhat is the phenotype?1. Normal male2. Color blind male3. Normal female4. Color blind femaleType number and press RETURN.No. Try again.$No. This offspring would be a normalfemale.$No. This offspring would be a normalmale.#No. This offspring would be a color blind male. Do these results seem to support#Mendel's theory of dominance (Y/N)?&Yes. This does appear to be an example!of simple Mendelian genetics. Butappearances can be deceiving.$No. This appears to be an example of%simple Mendelian genetics because all&the offspring of a homozygous dominant&parent and a recessive parent show the$dominant trait. But appearances can be deceiving.#Let's see what happens when a color%blind woman (X@Wc@ZX@Wc@Z) mates witha normal man (X@WC@ZY).Here is the Punnett square.$Only the females have normal vision!Why are the sons color blind?$Color blindness is dominant in males$They inherit their one gene for thistrait from their mother%Right. You have discovered one of the%characteristics of sex-linked traits.$Sons inherit these traits from their%mothers. Normal vision is still domi- nant, but the sons have only onegene -- a recessive one."No. Sons inherit sex-linked traits$from their mothers. Normal vision is&still dominant, but the sons have onlyone gene -- a recessive one.$Dominance is not universal, however.!Press @R to see what happens when!you cross these two snap-dragons.%All the offspring are pink! An artistmight think that the colors are&blended. But genes don't blend. A test"cross between two of the pink off-&spring might help explain what's goingon. Press @R to see the cross.$This cross shows that the genes must!be discrete, not blended: red andwhite both reappear in the F@2 generation.!What explains the pink offspring?#Alleles for both red and white pig-ments are expressedA third allele is expressed."No. Both red and white alleles for%Right. Both red and white alleles for#pigment color are expressed. If you&were to magnify the snap-dragon petal, you would see both pigments dis-$tinctly. But to the eye, they appear to blend.!Inheritance of pink color in snap#dragons is an example of INCOMPLETE&DOMINANCE. The genes for white and red%color are said to be CODOMINANT sinceboth genes are expressed.$Human blood type is also governed by%codominant genes. Actually, three al-%leles govern blood type in people: A, B, and O.&Each individual, however, has only two%genes for this trait. O is recessive.A and B are codominant.&What blood type might the offspring ofthese parents have?ABAOAABBB"No. A or B are possible. How aboutthe offspring of these parents?!Right. How about the offspring ofthese parents?&Right. Because A and B are codominant,both alleles are expressed.#No. Because A and B are codominant,both alleles are expressed.&Codominance, incomplete dominance, and%sex-linked traits are all examples of"the complex way in which genes are%passed from generation to generation.DETERMINING PARENTAL GENOTYPE"Offspring inherit genes for traits$from their parents. By analyzing the$frequency of different phenotypes in"a population of offspring, you can!determine the parents' genotypes.$Would you like to determine parentalgenotypes for: One trait Two traits%PURPOSE: To analyze the expression of$a trait in a population of fruit fly#offspring in order to determine the&genotypes of those offsprings' parents#METHOD: Group 16 computer-generated%offspring by phenotype. Use this data%to determine the offsprings' parents. ASSUMPTIONS:1. One allele for each trait is!dominant; the other is recessive."2. All offspring survive. Each flyrepresents about 20 offspring.!3. Each phenotypic group contains#equal numbers of males and females. PROCEDURE:1. Select trait.!2. Run the experiment to generate offspring.3. Count the offspring of eachphenotype and enter the number.!4. Select a hypothesis describingparental genotypes.&5. If you want, use the Punnett squareto test your hypothesis. 6. Verify your hypothesis. Body color Wing form Eye colorBrown (B) is dominant.Ebony (b) is recessive.Normal (N) is dominant.Dumpy (n) is recessive.Sepia (S) is dominant.Red (s) is recessive.How many offspring are brown?How many offspring are ebony?%How many offspring have normal wings?$How many offspring have dumpy wings?#How many offspring have sepia eyes?!How many offspring have red eyes?Select a hypothesis:support:1. Both parents are homozygous2. Both parents are recessive.!3. Both parents are heterozygous.$4. One parent is dominant, the otherrecessive, for this trait."5. One parent is heterozygous, theother dominant, for this trait."6. One parent is heterozygous, the other recessive, for this trait.&Would you like to use a Punnett squareto test this hypothesis (Y/N)?!Is your hypothesis correct (Y/N)? B: brownb: ebonyN: norm.n: dumpyS: sepias: red recessive. dominant. Yes, that is a valid hypothesis.Could any other combination of$parental genotypes account for theseoffspring (Y/N)? That's right. Congratulations onunderstanding basic Mendelian genetics.%You selected an incorrect hypothesis.$Wrong. All of the offspring show the&dominant trait, so at least one parent&must be homozygous dominant. The other&parent can be recessive, heterozygous,or homozygous dominant.$Wrong. All of the offspring show the%recessive trait, so both parents must be recessive.!Wrong. Half of the offspring show the recessive trait and half the%dominant trait, so one parent must be%recessive and the other heterozygous.#Wrong. One-quarter of the offspring$show the recessive trait, so both of!the parents must be heterozygous.%PURPOSE: To analyze the expression of'two traits in a population of fruit fly#offspring in order to determine the&genotypes of those offsprings' parents ASSUMPTIONS:1. One allele for each trait is!dominant; the other is recessive.2. The two traits assortindependently."3. All offspring survive. Each flyrepresents about 20 offspring.!4. Each phenotypic group contains#equal numbers of males and females. PROCEDURE:1. Select two traits.$How many flies have brown bodies and sepia eyes?$How many flies have ebony bodies and sepia eyes?$How many flies have brown bodies and red eyes?$How many flies have ebony bodies and red eyes?$How many flies have brown bodies and normal wings?$How many flies have ebony bodies and normal wings?$How many flies have brown bodies and dumpy wings?$How many flies have ebony bodies and dumpy wings?"How many flies have sepia eyes and normal wings? How many flies have red eyes and normal wings?"How many flies have sepia eyes and dumpy wings? How many flies have red eyes and dumpy wings?Your results fit which of thefollowing phenotypic ratios:9:3:3:11:1:1:13:11:1 All dominant All recessive This ratio suggests which of thefollowing sets of parents:#How many blocks represent offspring#that will show the dominant charac-teristic for both traits?$that will show the recessive charac-$teristic for the first trait and therecessive one for the second?dominant one for the second?'Enter number (0-16), then press RETURN.$Your responses suggest the followingphenotypic ratio:#Now cross the plants to see if thisratio is correct.!Do the cross results support your ratio (Y/N)?Yes. This ratio is correct.Wrong. This ratio is correct."That's right. The correct ratio isWrong. The correct ratio is!TRAITS: Lets you select trait(s).#MATES: Lets you select two mates tocross.$PUNNETT: Lets you complete a Punnettsquare.%CROSS: Lets you cross selected mates.EXIT: Returns to the main menu.#RUN: Lets you cross selected mates.Commands ProcedurePunnett square instructions$Theory \ FAHBody color/eye colorBody color/wing formEye color/wing form BbSs x BbSs BbSs x bbss bbSs X bbSs BbSs x BBss BBSS x BbSs bbss x bbss BbNn x BbNn BbNn x bbnn bbNn X bbNn BbNn x BBnn BBNN x BbNn bbnn x bbnn SsNn x SsNn SsNn x ssnn ssNn x ssNn SsNn x SSnn SSNN x SsNn ssnn x ssnn%Both parents are homozygous dominant.&Both parents are homozygous recessive.Both parents are heterozygous.!One parent is dominant, the otherrecessive, for this trait.One parent is heterozygous, theother dominant, for this trait.One parent is heterozygous, the other recessive, for this trait.To fill in a Punnett square: 1. Type first genotype and pressRETURN.!2. Use the arrow keys to move thecursor to the next cell.&3. Repeat steps 1 and 2 until you havecompleted the square.'4. You can use the arrow keys to return&to any cell that is incorrect. Reentergenotype and press RETURN.5. Press D when you are done. To interpret a phenotypic ratio, remember:#1. The first number represents off-!spring that are dominant for bothtraits.$2. The second number represents off-&spring that are dominant for the first#trait and recessive for the second.#3. The third number represents off-'spring that are recessive for the first"trait and dominant for the second.$4. The fourth number represents off-"spring that are recessive for bothtraits.htraits. and might!therefore trigger photosynthesis.%2. Extracted pigments absorb light ofall wavelengths.#3. My data do not support either ofthe above hypotheses.Check the data again. Plantsdo absorb red light.do not absorb green light.do not absorb all wavelengths."Do your data support either of theother hypotheses?Yes. Plants absorb red and blue$light, which trigger photosynthesis.Right. Your data do not supporteither of these hypotheses.%No. Plants absorb red and blue light,which trigger photosynthesis.$LIGHT: Lets you select a wavelength.$PURPOSE: To determine the absorptioncharacteristics of the separatepigments in a plant.#RATIONALE: The raw extract consists#of many pigments, including chloro-%phyll a, chlorophyll b, and accessory%pigments. We can determine the absor- bance characteristics of each ofthese.to photosynthesis.1. Each pigment absorbs red and blue light."2. The chlorophylls absorb red and blue light.$3. The accessory pigments (carotenes%and xanthophylls) absorb red and bluelight.Each pigment absorbs red and blue light.The chlorophylls absorb red and blue light.!The accessory pigments (carotenes%and xanthophylls) absorb red and bluelight.1. Select a pigment.2. Select a wavelength.3. Run the experiment. Read themeter.4. Record your results.$5. Repeat the experiment with a dif- ferent wavelength. Complete five%trials for good experimental results.%6. Run the experiment again with eachof the remaining pigments.PIGMENT Chlorophyll a Chlorophyll bCarotene XanthophyllYou will have better data from which to draw conclusions if you#try different wavelengths with thispigment. Do you still want tochange the pigment (Y/N)?"You must select both a pigment and#a wavelength before you can run the experiment.#No. Check your data again. Only the chlorophylls absorb red and bluelight.!Most of the pattern of absorbance%seen in the crude extract is actually%due to the absorption characteristicsof: 1. Carotenes2. Chlorophylls3. Xanthophylls$Right. Most of the pattern of absor-"bance seen in the crude extract is$actually due to the absorption char-acteristics of:Right.$No. Chlorophylls are responsible for!most of the pattern of absorbanceseen in the crude extract. They absorb light in the red and blueregions of the spectrum.#PIGMENT: Lets you select a pigment.m.#PIGMENT: Lets you select a pigment.rude extract. They absorb light in the red and blueregions of the spectrum.#Pigment: Lets you select a pigment.umber 2.#These results best support which ofthe following hypotheses:Right.#Now here's a thought to ponder: Why%does the rate of photosynthesis level off? Could you increase the rate%further by changing another variable?!This hypothesis is only partially correct. Greater light intensity increases rate of photosynthesisonly to a point, then the rate levels off.No. The second hypothesis is correct. Greater light intensity increases rate of photosynthesisonly to a point, then the rate levels off.No. Your hypothesis isProcedureSet Up: Lets you choose your instrument.Light: Lets you select a light intensity.!Table/Graph: Lets you go back andforth from a table to a graph"showing your experimental results. Run: Tells the computer to beginthe experiment.!Exit: Tells the computer that youare finished with the lab.#Press P to review the procedure forrunning this experiment.!Each mark on the manometer repre-sents 0.1 milliliter. When the"experiment begins, water fills the manometer up to the 1.0 ml mark.During the experiment, oxygen displaces some of this water. To measure how much, read the water"line at the end of the experiment.Record your answer with a digit!both before and after the decimalpoint, e.g., 0.1 > !"#$ % &'()*+,-./01211 1}1´1 1ěKNkşHMO HO-HGenetics: Disk 1G& HdP-H;<Review of MeiosisG‡h iWW>4N8˙A<Mendelian GeneticsG‡h iWW>48U˙A<Your Test Organism:GW The Fruit FlyG‡hiWW>4Ó8˙AjbcdKxHMO HMOHQśHQaCEWEWEWEWEWEe5f}dHM O HmOaFO2H QśHQ2H FĘH FaCEWEW EW EW EefšaM;OB<HN? GÍH? GN(H MĆO`ŒID FCa EW EWEôY IZ#\B<HNFÍHNFaWWWN9 Right.G8No. They have 23.G#\aM2OB<HM OG<HN? GÍHM OŇ<HN? GNaEWEWEôY IZ#\G<HNFŇHNFaWWWN9 Right.G8No. They have 4.G\dN HMlOa FO HC EqIEÔI!E HMAOqIFOÔI4O  H 11EW 1"98č˙q&H1EW 1"98č˙Ô&H1EW 1 "98č˙ 1aEWEWEWEWEWEefbM OaMFOqHM  OqH"EÔI#Eq HM-OÔI0OaCEWEW EW EW EW EefCdHM O HlOa FOTHQCaCEefPdHM Oa FOHFTHQaCEWEeaM FOf¤dHM OHZ OTHQHF HHM#OČCH CentromereGaMFOaC EWEWEWEe HMlOaM FOfădHM Oa FO HHM#OČCHK OHMZ OHFTHQNpCHM PpLHĺ˙ON(GHTetradGCaEWEW EW EW EW Ee HMlOaM FOfţdHM Oa FOp<HM  PpLHĺ˙O(GH2 OHMZ OHFTHQfAHMç˙O<H CentrioleGąKHOČFH Spindle fiberGCaEWEWEWEWEe HMlOaM FOfŔdHM Oa FOfAHç˙O<HF OąKHOČFHZ OHMZ OHFTHQCaEWEW EW EW EW Ee HMlOaM FOfƒdHM Oa FOHMZ OFHQHFCa EWEWEWEWEe HMlOaM FOfŔdHM Oa FO.H MżOHMZ OHF8HQ 1CaJEWEWEWEö˙1P I;< EW>3 8%A<EW>3ź 8A`ŒID F 1B6aJMFOaCEWEW EW EW E U3ô 6aJMFOaCEWEW EW EW E UÁdHM Oa FO.H MżOHMZ OHF9HQšHQCaEWö˙1(  H;<EW>3ý 80A<EW>3ˇ 8 A<EW>3ý 8A`ŒID FB?aMFOaCEWEWEe HMlOaM FOf3C DaMFOaC EW EW EW Ee HMlOaM FOf8aM FOaC EWEWEWEWCEWE UQdHM O 1a FO(H MĆO`ŒID F HMZ OHFHQšHQNCaEWEôY IZ \ 1 HMlOaMFO.H MżO2H QśHQ 9 aEWE8 a EW Eö˙1<  H;<EW>3Đ 8 A< EW>38A`ŒID FB.aMFOaC EWEe HMlOf3:3aMFOaCEWEWEe HMlOf”dHM O HMlOa FOŒHD F*HQ¨HQa CEWEWEWEWEWEe HMlOaM FOf†dHM O HMlOa FOŒHDFMZ OpHQa CEWEW Ee HMlOaM FOfśHM O HMlOa FOTHQ\HZ OHFTHQ\THQ\HZ OHFTHQ\HZ OHFTHQ\HZ OHFTHQ\HZ OHFFHQ\8HQ\HZ OH Metaphase IIG8HQšHQHZ OH FHQšHQHZ OHF~HM OaMFO.H MżOa CEö˙1<  H;<EW>38 A<EW>38A`ŒID FB‡ HMlOaMFOH QF<H Q~H Qś<H Qa CEWEWEWEWEe HMlOaM FOf3§‡ HMlOaMFOH QF<H Q~H Qś<H Qa C EWEWEWEWEe HMlOaM FOf—dHM O HMlOa FOHMZ OŒHDFpHQa CEWEWEWEWEe HMlOaM FOfň HMlOa FOTHQ\HFTHQ\THQ\HZ OHFTHQ\HZ OHFTHQ\HZ OHFTHQ\HZ OHF{dTH Q\HZ OĂdH FŇaHMí˙Oa CEWEWEW EW Ee HMlOaM FOfľHM OĂdHMF OŇaHMí˙OPH OaM FO-H MžOa C EW Eö˙1<  H;<EW>3h8 A<EW>3’8A`ŒID FB#aMFOa CEeaMFOf3ˇ#aMFOa CEeaMFOfNdaM FOa CEWEWEWEWE UHMZ O8LH Q\HMZ ObH Q\HMZ ObH Q\8HQaM FO%H MĎO`ŒID FaEWEôY IZ!!9$pHQaM 2OaEW E8*pHQaM FO EW EW EW EôY IZ!_aM FO!.9aEWEWE8 EWEe HMlOaM FOTH Q\f˘d HMlO HMlOa FOHQśHQ\bHQa CEWEWEWEWEWEe HMlOaM FOTH Q\f5ÂKNşHMO HO-HGenetics: Disk 1G& HFP-H;<Lab: Predicting GW OffspringG‡hiWW>4Ô/8s˙A<Mendelian GeneticsG‡h i>4ś8K˙A?5AjáKNşHMO HO-HGenetics: Disk 1G& HFP-H;<Lab: Determining GW Parental GenotypesG‡hiWW>4)t8i˙A<Your Test Organism:GW The Fruit FlyG‡hiWW>4˝h8,˙A?5AjŘcdK HMOxHMO HQaCEWEWEWEWEWE I'YZ"aM FOa" G. G I" &98Ea J E" 98!"„9Ea J E8 EaW EW Eefâd HMlOa FOTH QHCE HE H)EĽ H+E,H2E ,H-E8H EŞ8HEDHE´DHEPH*EšPHE\H Eź\H(EaC EWEWEWE UĐHM\ O HU O H\ OĽ Hj O,HU O ,HU O8HG OŞ8HG ODH@ O´DH9 OPH9 OšPH+ O\H+ Oź\H@ Oa FOŤTH!QHC1E H0E  H EĽ H4E ,H3E ,HE8HEŞ8H/EDHE´DHE PH,EšPHE \H Eź\HEef@d HMlOa FOH(QaCEWEWEWEef\d HMlOa FO*(H)Qb(H*Qš(H+QŇ(H/QaCEWEWEef(da FOaCEWEW EW EW EefŸd HMlOa FO*(H,Qb(H-Qš(H.QŇ(H2QaC EWEWEWEe HMmO*(H)Qb(H*Qš(H+QŇ(H/QfŃ HMlOaM FO*H)QbH*QšH+QŇH/Q*FH,QbFH-QšFH.QŇFH2Q.H MżOŒHD FaCEW( Iö˙1;<EW>318A<E>3 8ABÝ HMmOaM FOTH3Q8H1Qt-HM#OĹ HF HMđ˙O˛ H Os HFp Hâ˙OR HOuhHFrkHMö˙OaCEWEWEWEWEef3ő Ţd HMmOaM FOTH3Q8H1Qt-HM#OĹ HF HMđ˙O˛ H Os HFp Hâ˙OR HO}hHFzkHMö˙OaCEWEWEWEWEefœaM FO.#HM#O8#H2O8UHö˙OH*QDH+Q#H MÔO`ŒIDF 1aCEWEWEW EW!EZ$ 1]aM FOH.QDH-QaC$GREEN9#E8"EW$EW%Ee HMmOfĄd HMmOa FObH*QšH*QaCEWEWEWEWE U*FH*QbFH*QšFH*QŇFH*QaM FOaEWEW E U‚ HMmObH+QšH+Q\*FH+QbFH+QšFH+QŇFH+QaM FOaC EW EW EW E UŰ HMmObH*QšH+QNHP1G\*FH*QbFH*QšFH*QŇFH*QKHF1GaM FO.H MżO`ŒID FaCEWEW( Iö˙1;<EW>3ö#8A<E>3Ż$8AB˛aM FONHOKHO_HM$$P_Hě˙O HDOMINANTG—HM$$PťHOÖH RECESSIVEGaCEWEWEWEWEef3c%˛aM FONHOKHO_HM$$P_Hě˙O HDOMINANTG—HM$$PťHOÖH RECESSIVEGaCEWEWEWEWEefˆd HMmOa FO~H0QN…HTtGN 1#H MÔO`ŒIDFaCEWEWEWEWE IZ%L 1aM FO~H*QaC%TALL%T /9E8EW EW Eefvd HMmOaM FObH*QšH+Q*1H*Qb1H*Qš1H*QŇ1H*QaC EW EW EefLaM FODHF 8HFCaEWEWEWEWEWEefod0HM2O 8HFb4H0QNi?HTtGNš4H0QNĄ?HTtGNaM FO.H MżO`ŒID F 1*RH0QN1]H? GNbRH0QNi]H? GNšRH0QNĄ]H? GNŇRH0QNŮ]H? GN XH FaCEWEWö˙1( I;<EW>3r(8'A<EW>3r(8A<EW>3Ď(8A 1BVaM FO*RH*QbRH*QšRH*QŇRH+QaCEWEW EW E U3')VaM FO*RH*QbRH*QšRH*QŇRH+QaCEWEW EW E UJaM FObH0QNiHTTGNšH0QNĄHttGNíPHM&O*RH0QN1]HTTGNbRH0QNi]HTtGNšRH0QNĄ]HTtGNŇRH0QNŮ]HttGNaM FOaEWEWEe HMmOaM FObH*QšH+Q HFfěd HMlOaM FOb)H0QNi5HTTGN5HGenotypeG]1HMú˙OW1HO@JOš)H*Qş1HOŔ1HO@Jú˙OČ5H PhenotypeGaC#EWEWEWEe HMlOfŔaM FO HlO.H MżO`ŒID F~H0QN…HTtGNaCEWEö˙1( I;<EW>\~H*QaWWWE8&A<EW>\~H*QaWWWE8ABŔaM FO HlO.H MżO`ŒID F~H0QN…HTTGNaCEWEö˙1( I;<EW>\~H*QaWWWE8&A<EW>\~H*QaWWWE8ABd HMlOa FO 7H0QN#BHTTGNX7H*Q%"H#F* HM9O*IOcIOš7H0QNĄBHTtGNŇ7H*Qš"H"F¤ HM?O¤IOăIOaCEWEW EW!EW"EefKd HMlOa FOb(H)Qš(H+QaCEWEWEWEef*aM FOaCEWEWEWE UQd*RH*QbRH*QšRH*QŇRH*QaM FOaC EW EW E[\(aM FO9 E8 EWEWEef\daM FOaCEWEWEWEWEe HMlOa FOb(H)Qš(H+QfĽ HMlOa FOH(QaCEWEWEWEWEWEe HMlOa FOb(H)Qš(H+Q*RH*QbRH*QšRH*QŇRH*Qf5tcdK HMO2HLAB: PREDICTING OFFSPRINGGWW I#CEWEWEWEWEWEWWEW E[94†ScdK HMO.H MżOH#C EW EW Jö˙12I;< EW J> 1,3Ü08$A< E> 1,3E8A`ŒID FBrcdK HMO H#CEWEWE. H MżO  HEWEWEWEWEWEWEefxdK HMO H#CEWWEW JEW JEW JEW JEW JEW J EW!Eef K1)4é=űNaMdOH O HMO H"Cö˙g;<E h>4Ň28¨˙A I<E h>458ˆ˙A I<E h>4x78h˙A I<E h>4Í>8H˙A I<E h>4ÎB8(˙A I<E h>3NB8˙AjZHMśhO HMvO HMShO HMp P# H MĐOŒHD"FŒ HD"F%C1&1'  H HMd O&ET 98N 9&1&&91&8' 9& 1&&&91&88Ľ˙8…˙aMdOHH M‡O`ŒID" Fa%C&9Tq*+lEWE&9Aq*)lEWE&9Yq-.lEW E&9Rq.,l EW EUNaMdO HMvOMX HşjP:HDFemale:G:HHDMale:G1)5ŒZHMśhOaMdO)*:eaYou must select a trait beforeGWyou can select mates.GHH M‡O`ŒID" FU5×n: HMlO@HH O"H OVH O< HMf P"14x6'1(V14x6@HH&&' B F: HMlOX HMşjP@H&&( B F(r's1)5ć# H MĐOŒHD"FŒ HD"F&C1'  HDHMZ O&' EH'9o8oT :N 9'1''91'8' 9' 1''&91'88¨˙8k˙# HMĐ(O5vƒHM<OZHMśhOaMdO)*:=a'CEWEWEHH M‡O`ŒID" FU5v”+tuaMdO H MúOŒ˘HD" FŒŹHD" FŒśHD" Fyz.9œaMdOaMdO  HMP Iindicates wrong gameteG H MúOŒ˘HD" FŒŹHD" FŒśHD" F{8Z˙aMdO;aMdO H MúOŒ˘HD" FŒŹHD" FŒśHD" Fyz.9ßaMdOaMdO  HMP Iindicates wrong genotypeG  HMP Iindicates wrong letter orderG H MúOŒ˘HD" FŒŹHD" FŒśHD" F{8˙laMdO(C#E H$E  H;< I%E H<&EŞ H<'E H<(E ?Aö˙1,´H MŔOśHD FB('/ 0:ş('0 0:Ł('0 0:Œ('0('0/ 0:gaMdOHH M‡OaThat is not the right ratio.GW Try again.GŒHD" FU8”ţ+KMHO  HPhenotypic ratio:G JMO  H $ E)C  HEWE H;<E HE H<E HE H<E HE H E H< E H E H E ?A,´H MŔOśHD FB 1*1) HMžOHMO H)CYOUR HYPOTHESIS:GWWI*9 EWE*9 EWE*9EWEW E*9 EW EW EaNow cross the plants to see Gif yourGWhypothesis is correct.GHH M‡O`ŒID" FU3é=âKN:HDFemale:G:HHDMale:G HMOHOX HşjP)&9o@H&&( B Fn"H(9o8o@HH&&' B FVH'9o8o)&94[?5Œ)*9 4[?3§?8yaYou must select PUNNETT G before you GWcan carry out a cross.GHH M‡O`ŒID" FUaMdO5JnZHMˇhO1b " Hp 1:Ď˙5Ł1)a)C*9 EWE*9 EWE*9EWEW E*9 EW EW E HDo these results support GWyour hypothesis (Y/N)?G I[o('/*0:X('0*0:A('0*0:*('0('0/*0:aMdO.9 Right. G8 Wrong. GThe correct conclusion is:G H('/9 EWE8^('09 EWE8B('/9EWEW E8!('09 EW EW E8HH M‡O`ŒID" FUZHMˇhO5aMdO9 Right. G8 Wrong. GYour original hypothesisGW was correct.GHH M‡O`ŒID" FUZHMˇhO5~)9ta You have not Gcompleted the entireGWlab sequence. Are you sureG that youGWwant to exit (Y/N)? G[.93Ő15ůž HMT#O HMP,P;  H<CommandsGLh>4ÉC,93é=83*QA H< ProcedureGLh>4QD,93é=83*QA H<PunnettGLh>4E†,93é=84ě†3*QA H<TheoryGLh>4ś,93é=83*QAj†K HMO H<CEW JEWEW JEWEW JEW JEHH M‡O`ŒID" FU5˛K HMO H#CEWW,9 $F8EW JEW JEW JEW JEW JEW J EW!EHH M‡O`ŒID" FU5vcdK HMO H$CEWEWE. H MżO  H#CEWEWEWEWEWEWEefAdK HMO H7CEWWEWEW JEW EefxdK HMO H#CEWWEW JEW JEW JEW JEW JEW J EW!Eef K1)4*QűNaMdOH O HMO H"Cö˙g;<E h>4BG8¨˙A I<E h>4ĽI8ˆ˙A I<E h>4ŘL8h˙A I<E h>4ÔQ8H˙A I<E h>4ÎB8(˙A I<E h>3NB8˙Aj›MaFO>HM×hO HM¨O HM7jO HM˘ P# H MĐOŒHD"FŒ HD"F%C1&‹ HM– O&E/ G&ET 98N 9&1&&91&8' 9& 1&&&91&88Ľ˙8u˙ŕaMFOHH M‡O`ŒID" Fa%C&9PTAq*)l+/m 1/NEF IEWEW IE IEWE8MYRq-2l.,m:1/NE IEW EW I E I EW EUUaMON HM¨OM< HÚjPHHM:GHF:G1)5‹aMFO>HM×hO)*:daYou must select traits beforeGWyou can select mates.GHH M‡O`ŒID" FU5ën: HMŸO@HH OH$ O"H8!OHH$ OVH8!O< HMš P"14K'1(12rV14K12s: HMŸOM< HÚjPH&/( A F GHH&/' A F G1)5b# H MĐOŒHD"FŒ HD"F&C1'  HDHM O/' EH'9o8$'9o8' 9o8o'&9118'911811''/9128''/912812T :N 9'1''9 1'8' 9' 1'' &91'88¨˙8ďţ# HMĐ(O5;H09o8$09o80 9o8o5šƒHM<O>HM×hOaMdO)*:aa'CYou must select two traits Gand twoGWEWEHH M‡O`ŒID" FU5TvFtuaMdO H MúOŒ˘HD" FŒŹHD" FŒśHD" Fyz.9ŕaMdOaMdO  HMP Iindicates wrong gameteG  HMP Iindicates wrong letter GorderG H MúOŒ˘HD" FŒŹHD" FŒśHD" F{8˙aMFO>aMdO H MúOŒ˘HD" FŒŹHD" FŒśHD" Fyz.9âaMdOaMdO  HMP Iindicates wrong genotypeG  HMP Iindicates wrong letter GorderG H MúOŒ˘HD" FŒŹHD" FŒśHD" F{8˙Á;CaMdO`J MúO JEaEWEWE I€aM(OEWEWE I€aM-OEWEWEWE I€aM-OEWEWEWE I€[MadO EW E‚ IWI  ~HH M‡O`ŒID" FU1)5¨KNHF:GHHM:G HMOHO< HÚjP)&9AnH&/( A F(10"14›LHH&/' A F'10V14›L)&94mR5—)*9 4mR3šR8„aMdOYou must select PUNNETT G before you GWcan carry out a cross.GHH M‡O`ŒID" FUaMdO5JnM>HÖhO1Z " Hp 1:Ď˙5…1);Ca YOUR RATIO: GW EWE I[MadO9$‚9E8E I. G8!‚9E8E I. GDHH M‡O`ŒID" FUM>HÖhOadO5vdK HMOxHOFHMx8O  HM6O HtOCaEWEWEe5fXdaMFOaEWEWEWEWE1UNs#HT G \* HT GNefdaMFOa EW EW EW EW EUNN5HT G \ Ht GNeNN4H G H Gs#H G* H GNfĘdaMFOaEWEWEWEUNn5HT G \* HT G \v Ht G \* Ht GNeNn5H G* H Gv H G* H GN4H G H GNfŞdaMFOaEWEUNv5HT G \* HT G \n HT G \* HT GNeNn5H G* H Gn H G* H GNffdaMFOaEWEWEeN1v5H O* H On H O* H ONf(daMFOaEWEWEWEWEefIdaMFOMNj3H@ PB  HO HN HomozygousGNjEH@ PB  HO HN HeterozygousGaEWEWEWEeMNj3H@ PB  HO HMNP OMNjEH@ PB  HO HNMP OfüCaMFOa EW!EW;ö˙1F I<"EW>A<#E>A*H MÄO`ŒID FBM HlOaFO*R HMOxHO11Z&:(Sć9 18â˙81" 18Ď˙a 9$E8(EW%EW&EW'EYeM HlOFHMx8O  HM6O HtONs#HT G* HT GN5HT G Ht Gn5HTTG* HTTGn HTtG* HTtGNMNj3H@ PB  HO HN HomozygousGNjEH@ PB  HO HN HeterozygousGf3śKdC HMOHOX HşjPH Female: TtG H*QHHMale: ttG H+Q13Tqvrs”+tuaEWE[.93aFaMFOEWEH MÖO`HFWFżyaMFOz9%za9Good. G88  EW E8)za9  EW E8EWEWEWEW EWE[aMFO913vrsaMFOH Female: tt G H+QHHMale: TTG H*QuaEWEH MÖO`HFWFyaMFOz9%za9Good. G8l8 $EW%E8$za9 "EW#E8&EW'EW(E$\aMFOwxa8aMFOwwwxaĂEWE H MúOŒ˘HD" FŒŹHD" FŒśHD" FyzaMdO9 Very Good.GW8N8K  HMP Iindicates wrong genotypeG  HMP Iindicates wrong letter GorderGWEWE[aMdO9Ľ H MúOŒ˘HD" FŒŹHD" FŒśHD" F{zaMdO9 Very good.GW8B83EWEWEWEWEUaMdOwxa8 wxaBH MĘO`HEnter number from 0 to 4 andGW press RETURN.GaEWEY IZaMdOM39J9EW8-¨=H P¨ H PaNo.GWEW E8w9EW8H¨=H P I P¨ H P I PaEWfor this cross is 1:0.Gž$\M HoO1139CZ&:7*Q 1+Qć9 18Ó˙81" 18Ŕ˙84Z&:+*Qć9 18ß˙81" 18Ě˙efoKd C HMOHOT(H*Q8 I*QT( H TtAa TtAaGaEWEWEWEef-daMFOaEWEWEWEW EW EefdaMFOa EW EW Eef  CaMFO HlOM< HÚjPHF: TtAaG H*QHHM: TtAaG HSTAqvrsFtuaEWEWEŸH MĘO`HFWF|}yzaMFO9 EWE8;EWEWEWEUwxaMFOEWEWE¨H MÂO`HUse   to move cursor, typeGWF|}yzaMFO9E81EWEWEWEUwxaMFOE4–,f 2 HNZ4NaMFO4TtAa9 Good. G8MEWEWEWEU,f 2 HNTtAa GNaMFO E„,f 2 HNZ4N4Ttaa9WaMFOEWEWEWEU,f 2 HNTtaa GNM ONd CwxaMdO`J MúO J;Fa!EW"EW#EW$E:Y IZ 9!aMFO(EW)EW*EW+EefUaMdO`J MúO J;Fa 9%EW&EW'E8 !CEWE?d!CY IZ9!aMFO EW EW EW EefLaMdO`J MúO J;Fa9 EWE8 EWEPdY IZ96aMFO EWEWEWEeaMFOEWEfQaMdO`J MúO J;Fa9EWEWE8 EWERdY IZaMFO9E8EWEWEeaMFOEWEfxM HoOaMFO11ć&988*Q 18â˙1" 1*Q 1)Q 1)Q 1)Q 1+Q 1+Q 1+Q 1/Q( 1xIDPHENOTYPIC RATIOGx HD9:3:3:1GaEWEWEWEeM HoOaMFOM< HÚjPHF: TtAaG H*QHHM: TtAaG HSuwxaEWEf5…cdK HMOxHOTH4QaCEWEWEbdINCEN IE( HM´O…NbM OŠdN HMOaM FOHQ UH DrosophilaGW melanogasterG`HMBOkHXOaCEWEWEaŽ HMBOWIXOefSd HMlOaIFOHCE  HM;O#H1.GW2.GW3.GW4.GW5.GW6.G#HEWEWEWEWEWE7HM$OĽHEĽ HMgOĽ#H EW EW EW EW EWEaCEWEW Ee HMlOa FOTH4QHQ UH DrosophilaGW melanogasterGfÉaM FO)H MĆO`ŒID FaC EW EW EY IZ#\CNDHEĽI EN\aM FOa#9EWEWEWEWE8EWEWEWEWEWEefÂd HMlOa FO#R HS\ŕXHS\b2HS¨HS\îHSZHSF(HS<HS\<PHS UHS}HSŕ7HSaCEWEWEW IEW\EW\Eef  1 HMjOa FO*#HQš-HQ7HFemaleG5 HM-OŞHMaleG¨ H!O,H MÂO`ŒID FaCEWEWEW( Iö˙1;<MaleGW>aM FOaEWE8)A<FemaleG>aM FOaEWE8ABefŻ 1aM FOF;H;Qś=H<Q,H MÂO`ŒID FaCEW( Iö˙1;<MaleGW>aM FOa EWE8)A<FemaleG>aM FOa EW E8ABef/ 1aM FO HMjO*#HQš-HQTH9QÄH:Q2HFemaleG0 HM-OŞHMaleG¨ H!O,H MÂO`ŒID FaCEW EW EW( Iö˙1;<MaleGW>aM FOaEWEWE8.A<FemaleG>aM FOaEWEWE8ABefŻd HMjOa FO*#HQš-HQF;H;Qś=H<Q7HFemaleG5 HM-OŞHMaleG¨ H!OaCEWEWEWEWEWEef?d HMjOa OOaCEWEWE…N93qÎ HMlOa FOT#HQp;H;Q~H9Q.H MżO`ŒID FaCEWö˙1( I;<EW>+aM FOa EW EW EW EWEef8&A<E>aM FOaEW Eef8ABŐ93gr HMlOa FOT-HQp=H<Q~H:Q.H MżO`ŒID FaCEWö˙1( I;<EW>aM FOaEW Eef80A<E>&aM FOaEWEWEWEef8AB‰daM FO HMjO*#HQš#HQFH>QśH?QaCEWEWEWEWEWEe HMlOa FOfad HMjOa FO*#HQš#HQTH@QÄHAQaCEWEW EW EefUd HMjOa FO*#HQš#HQaCEWEWEWEWEWEef;d HMjOa FOFH=QaC EW EW Eef;d HMjOa FO HQaCEWEWEef5ŔcdK HMO.H MżOH2CLAB: GEWW IEWEWEWEWEWEWEW Jö˙12I;< EW J> 1,3ët8)A< E> 1,3Ą‡8A`ŒID Fö˙1BbdK HMO H3CEWEWEWE. H MżO  HEWEWEWefRdK HMO H3CEWW EW EW J EW EW J EWEef…dK 1 HMO H3CEWWEW JEWEW JEWEW JEWEW JEWEW JE 1ef K1)4Ä˝NaMdOH O HMO-H"Cö˙g;<E h>4ôv8¨˙A I<RUNG h>4Ňx8†˙A I<E h>4„8f˙A I<E h>3eƒ8F˙AjHMThOZHMśhO HMvO HMp P# H MĐOŒHD"FŒ HD"F4C1&1'  H HMd O&ET 98N 9&1&&91&8' 9& 1&&&91&88Ľ˙8˙ĆaMdO&9Bq""l&9Nq""l&9Sq""lNa4C&9 EWE&9 EWE&9 EW EHH M‡OŒHD" FU1)4Ä5‰)*93Šy8yaYou must select a trait Gbefore you can GWcarry out a cross.GHH M‡O`ŒID" FUaMdO5JnZHMśhO1b " H 1:Ď˙5-„ 11+4߀PHMP'O4]y1)ą H MO`ŒID;Fa4C&9 E&9 E&9EWYZ. \W&9 E&9 E&9EWYZ- \.- 94EaMFO8O˙ KMHO HDominant offspring: G. GWRecessive offspring: G- G 14CW JEWö˙1  H;<EW&EW J<EW J<EW J<EWEW J<EWEW J<EWE ?A,´H MŔOśHD FB 1* 14ÄÁACaYOUR HYPOTHESIS:GW*9E*9E*9E*9 EWE*9 EWE*9 EW E4CW JEWE I[9$aMFOZHMˇgO4l81Ć++/+/9**/*/18+*1aMFOFH MO`ŒID" Fa999 5F8#Your hypothesis was G correct. G89 6FW8+Right. That hypothesis is G incorrect.GW+9&The parents were both G recessive.G+9)The parents were both G heterozygous.G+9=One parent was G heterozygousGWand the other recessive.G++/+/9Both parents were dominant.GUaMFOa5CEWEWE I[aMFOFH MO`ŒID" Fa++/+/9?9EWEWE8&6C 1EWEWEWEWE 18h9Q6C+9EWEW E+9EWEWE+9 EW EW EW E85CEWEWEU4Ä5}aMFOHH M‡OŒHD" FaThe total number of offspring GmustGWequal 16. Please reenter.GU5KN HMOHOX HşjP)*9>HTRAIT: GW4&FW J4& FW4& FW)*9+14߀4]yPHMP'P\HP P*\HO*RHD OFFSPRINGG^HDDom.G>^HDRec.GkHD. G>kHD- G5‹9rs9rs9rs9rs9rs9rs5U*14߀v”+tuaMdOwx H MúOŒ˘HD" FŒŹHD" FŒśHD" Fyz:ÓaMdO  HMP Iindicates wrong genotypeG  HMP Iindicates wrong letter orderG H MúOŒ˘HD" FŒŹHD" FŒśHD" F{8$˙aMdO aYOUR HYPOTHESIS:GWAC*9E*9E*9E*9 EWE*9 EWE*9 EW EW J4F I[91815Ť)9ƒa You have not Gcompleted the entireGWlab sequence. Are you sureG that youGWwant to exit (Y/N)? G[.9,935v83‰1)1&1'1.1-1,5ůŒ HMT#O HMP,P;  H<CommandsGLh>4 …,93Ä83zA H< ProcedureGLh>4„…,93Ä83zA H<PunnettGLh>4E†,9 4ě†3z83ÄA H<TheoryGLh>4˝h,93Ä83zAjuK HMOFH MO`ŒID" F H<CEW J,9E8EW JEWU5żK 1 HMOFH MO`ŒID" F H3CEWW,9E87FW JEWEW JEWEW JEWEW JEWEW JEU 15ĽK 1 HMOFH MO`ŒID" F HBCEWWEWEW JEWEW JEWEW JEW EW EW J EU 15łK 1 HMOFH MO`ŒID" F HBC EW EW JEWEWEW JEWEWEW JEWEWEW JEWEWEU 15gcdK HMO H7CEWEWEWE. H MżO  H3CEWEWEefcdK HMO H7CEWWEWEW JEW EW J EW EW J EW EefˆdK 1 HMO H3CEWW7FW JEWEW JEWEW JEWEW JEWEW JE 1ef K1)4z˝NaMdOH O HMO-H"Cö˙g;<E h>4É8¨˙A I<RUNG h>4ţ‹8†˙A I<E h>4„8f˙A I<E h>3eƒ8F˙AjHM>hO HMO HM P# H MĐOŒHD"FŒ HD"F=C1&1'  H HM‰ O&ET 98N 9&1&&91&8' 9& 1&&&91&88Ľ˙8˙5aMdO&9BSq""l&9BNq""l&9SNq""lNa4C&9-E IEWEW IE IEW E&9-E IEWEW IE IEWE&9-E IEW EW IE IEWEHH M‡OŒHD" FU1)4z5‰)*93‰Œ8yaYou must select a trait Gbefore you can GWcarry out a cross.GHH M‡O`ŒID" FUaMdO5„ 11+4’4]”1)î1  H MO`ŒID;Fa&7 CEWE I€ \  1  &:–aM2OaEWE I€ \  1  &:aaM2OaEWE I€ \  1  &:,aM2OaEWE I€ \  1 ” :‹aMFOHH M‡OŒHD" FaThe total number of offspring GmustGWequal 16. Please reenter.GUaMFO3ŚŒčKMHO HYOUR PHENOTYPIC RATIO:G I‚:CW  HEWEö˙1 H&= C;<EW J<EW J<EW J<EW J<EW J<EW ?A,´H MŔOśHD FB 1*4z~aYOUR HYPOTHESIS:GWThe parents are G&= C*E. G4CW JEWE I[9 aMFOZHˇgO4˘’81â+*1aMFOFH MO`ŒID" Fa9?9 5F8)Wrong. Your hypothesis is G correct. G8a9 6FW8+Right. That hypothesis is G incorrect.GWThe parents were G&= +F. GU™KN4C HMOHO@ HÖjP)*9?HTRAIT: GW&9[BodyGWJMO JEW EW JEyeGWJMO J#EW$E&9\BodyGWJMO JEW EW JWingGWJMO J!EW"E&9[EyeGWJMO J#EW$EW JWingGWJMO J!EW"E)*9 +14’4]”5‹9rs9rs9rs9rs9rs9rs5U*14’vJtuaMdOwx H MúOŒ˘HD" FŒŹHD" FŒśHD" Fyz:ÓaMdO  HMP Iindicates wrong genotypeG  HMP Iindicates wrong letter orderG H MúOŒ˘HD" FŒŹHD" FŒśHD" F{8$˙aMdObaYOUR HYPOTHESIS:GW&= CThe parents are G*E. G4CW JE I[91815JnZHMśhO1b " H 1:Ď˙5ike to see this GW set-up (Y/N)?G[94^Ÿ 1ŹfK 1 HMO.H MżO H-CEWE H<Cö˙1;<EWEWW> 1"8?A<EWEWW> 1"8'A<EWE> 1"8AŒHD FBžfK 1 HMO H5CEW J=CEW JEW JEWEW JEWEW JEW EW J EW EW EW EWEWEghĚK1 NM‚HOU HÂrO H2QZ"HM7IONx$H>$Fp3H>%FvBH>&FrQH>'F^`H>(FNZsHM7 ON\sHsecondsGeH H”˘Na JM <OH O HMOZH.Cö˙i;<E j>3›—8Ą˙A I<E j>4… 8˙A I<E j>3Ł™8a˙AlƒHM<O 9ĺa J=CEWEWE[9ÂK‚HMOU HMÂrO H2QZ"HM7IONx$H>$Fp3H>%FvBH>&FrQH>'F^`H>(FNZsHM7 ON\sHsecondsG1 3÷–83 ›eH H”\HMO 1eH0 GRH MwO`ŒID0Fa J9C!EW"EW#EW$EUƒHM<O ‘ š H’NM ™ sH O š sHN•  1  H2Q^HMOeH0 G3÷–e 9÷a J=CEWEWE[9ÔK HMO‚HMOU HMÂrO H2QZ"HM7IONx$H>$Fp3H>%FvBH>&FrQH>'F^`H>(FNZsHM7 ON\sHsecondsG1 3÷–83 › 9)a J1CEWEWE[93N83÷–a J1C EW EW EWE[9 1 3N83÷–˛KfdHMO'HMÂbO,HM7IONJH>$FBH>%FF%H>&FB4H>'F0CH>(FN,VHM7 ON.VHsecondsGN1 9QNp H’p VHNM Op VHN• 18§˙2H MľO`ŒID F 1aJ<CYOUR HYPOTHESIS:GW J"9  EW EW"9 EWE"9JEWEW›H2CE[ 1ÂK9"93Ež83‚8 "93‚ HMO2H MżO`ŒID F H>CEWEW J<Cö˙1;<"9EWEWW8  EW E>3‚8A<EWE>3Ež8ABÁK HMO2H MżO`ŒID F H>CEWEWEWEW Jö˙1;<EWEW EW J>3ţž8?A< EW EW J>3*Ÿ8#A< EW EWEW J>3*Ÿ8ABˇK HMO2H MżO`ŒID F H>CEWEW Jö˙1;<EWEW EW J>3ţž8?A< EW EW J>3*Ÿ8#A< EW EWEW J>3*Ÿ8AB*K>C HMOHEWE3Qą/K>C HMOHEWEWE3Qą5%K HMO‚HO H2QZH: FXHMń˙Od7HALGAEGb<Hö˙OdsHMETHANOLGbxHö˙Oš HMrrOž"HM7KONÜ$H>$FÔ3H>%FÚBH>&FÖQH>'FÂ`H>(FNJH M‚O`ŒID0FU5on HMTO HMPP;  H< ProcedureGLj>4:˘A H<TheoryGLj>4 vAlBKNM HO‚HOU HÂrOZ"HM7IO H2Q\HMOeH0 GNx$H>$Fp3H>%FvBH>&FrQH>'F^`H>(FNZsHM7 ON\sHsecondsG1 9QNš H’™ sHNM Oš sHN• 18§˙5§KN HMOFH MO`ŒID0F H-C#EW J=CEW JEW JEWEW JEWEW JEW EW JU5efK HMO2HLAB: CALVIN'S EXPERIMENTGW APPLIED TO G, F PLANTSGWW I9CEWEWEWEWEWEW EW EghśfK 1 HMO H9C EW EW E( H MżO  HEWEWE. H MżO  HWould you like to see this GW set-up (Y/N)?G[94ľ­jfK 19C HMOHEWEW JEWEWEW JEWEWEWEWE 1gh›fK HMO.H MżO H-CEWE H9Cö˙1;<EWEWEWW> 1"8.A<EWEW EWW> 1"8AŒHD FBžfK 1 HMO H5CEW J=CEW JEW JEWEW JEWEW JEW EW J EW EW EW EWEWEghÚK1 NM‚HOU HÂrO H2QZ"HM7IONb$H:#Fv0H:$Fr<H:%FwHH:&FrTH:'F_`H:(FNZsHM7 ON\sHsecondsGeH H”˘Na JM <OH O HMOZH.Cö˙i;<E j>3S§8Ą˙A I<E j>4Ż8˙A I<E j>3sŠ8a˙AlƒHM<O 9ýa J=CEWEWE[9ÚK‚HMO‚HOU HÂrO H2QZ"HM7IONb$H:#Fv0H:$Fr<H:%FwHH:&FrTH:'F_`H:(FNZsHM7 ON\sHsecondsG1 3ŻŚ83ĘŞ 1eH H”\HMOeH0 GRH MwO`ŒID0Fa J9C!EW"EW#EW$EUƒHM<O ‘ š H“NM ™ sH O š sHN•  1  H2Q^HMOeH0 G3ŻŚU 9ça J=CEWEWE[9ÄK‚HMOS HÂrO H2QZ"HM7IOb$H:#Fv0H:$Fr<H:%FwHH:&FrTH:'F_`H:(FZsHM7 O\sHsecondsGN1 3ŻŚ83ĘŞ 9)a J1CEWEWE[93N83ŻŚa J1C EW EW EWE[9 1 3N83ŻŚŹKfdHMO'HMÂbO,HM7ION4H:#FHH:$FDH:%FJ+H:&FE7H:'F1CH:(FN,VHM7 ON.VHsecondsGN1 9QNp H“p VHNM Op VHN• 18§˙2H MľO`ŒID F 1aJ:CYOUR HYPOTHESIS:GW J"9EWEWEW"9 EWEW›H2CE[ 1tK"098d."098T HMO2H MżO`ŒID F H:CEWEW J3B­RK HMO2H MżO`ŒID F H:CEWEW J7ö˙1;<EW>3˜­8"A<EW>3{­8A<EW>3˜­8ABdH:C EW Egh3’˛dH:C EW Egh3’˛JK HMO‚HO H2QZH: FXHMń˙Od7H CORN LEAFGdAHDISKSGb<Hö˙OdsHMETHANOLGš HMrrOž"HM7IONĆ$H:#FÚ0H:$FÖ<H:%FŰHH:&FÖTH:'FĂ`H:(FNbxHMö˙OJH M‚O`ŒID0FU5od HMTO HMPP;  H< ProcedureGLj>4¨°A H<TheoryGLj>4ՅAlÓKNM HO‚HOU HÂrO H2QZ"HM7IONb$H:#Fv0H:$Fr<H:%FwHH:&FrTH:'F_`H:(FNZsHM7 ON\sHsecondsG_1 9QNš H“™ sHNM Oš sHN• 18§˙5§KN HMOFH MO`ŒID0F H-C#EW J=CEW JEW JEWEW JEWEW JEW EW JU5  1sHMOH MćOH>CEWEWE 1c"2HEWEWEWEWEW1 1 1 4`˛1 4`˛1 4`˛1 4`˛1 4`˛HM#OxH>C 9EW EW!E8)"EW#E–2H&EW$EW%EW'EW(EWg1 h502Hd 9  1 D . G  1 5OK´1 1 HMO~HMOH MćOH>CEWEWE 1Hc?CEWE"2H:C#EW$EW%EW&EW'EW(EW1 1 1 4ćł1 4ćł1 4ćł1 4ćł1 4ćł1 4ćłHM#OƒH>C 9EW EW!E81"EW#E:C–2H$EW#EW&EW%EW'EW(Eg1 h502Hd 9  1 D . G  1 5502Hd 9  1 D . G  1 51 5