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Prophase I Metaphase I Anaphase I Telophase I Metaphase I Metaphase IIMaleFemale Polar bodyEggSperm Anaphase II Fruit fly Interphase I Telophase IITTttTt?P@1F@1F@2F@3 Heterozygous HomozygousDominant Recessive PhenotypeGenotypeUse @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 across#the top of the square. Press RETURN#to see the gametes for a homozygoustall (TT) female pea plant.%They plot all of the possible gametes$of the male parent along the side of#the square. Press RETURN to see the$gametes for a heterozygous tall male pea plant.$To fill in the Punnett square, gene-%ticists carry the male gametes across!all of the boxes. Press RETURN tosee.#They carry the female gametes down.Press RETURN to see.$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$gametes go down the side. They are tand t.#No. The female gametes go along thetop. They are T and t. No. The male gametes go down theside. They are t and t.%Do you want to try filling in anotherset of gametes (Y/N)?!Good. Now fill in each box of the#Punnett square. Press D when done."The highlighted squares are incor- rect. Remember to bring the male#gametes across all the boxes. Bring%the female gametes down. Press RETURNto see.!Do you want to try filling in theoffspring again (Y/N)?"Very good. How many of the squaresrepresent tall offspring?Right.No. 00 squares represent tall offspring.#How many represent short offspring?No. 00 squares represent short$The ratio of tall to short offspringfor this cross is 0:0.#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 RETURN 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 RETURN tosee the correct genotype.Now try this one.$Here's the completed Punnett square.%Count the number of boxes that repre-sent tall offspring with axialflowers.%Good. Now count the boxes that repre-!sent tall offspring with terminalflowers.%No. Tall offspring with axial flowers can have either of the following#genotypes: TTAA or TtAa. Nine boxescontain these genotypes."Now count the boxes that represent%tall offspring with terminal flowers.%Right. How many boxes represent shortoffspring with axial flowers?#Right. And how many boxes representshort offspring with terminalflowers? Very good. No. Tall offspring with terminalflowers can have either of the"following genotypes: TTaa of Ttaa.$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.TRAITSMATESRUNCROSSEXIT?!Press @U @D to see choices. PressRETURN to select.Press @R to continue.%Enter a number from 0 - 4, then pressRETURN.&Enter a number from 0 - 16, then pressRETURN.$Enter number and press RETURN. Press D when done.LAB: PUNNETT SQUARES You can predict the genotype and"phenotype of offspring if you know%the parents' genotypes. This lab lets"you use a special tool, called thePUNNETT SQUARE, to make these predictions."Do you want instructions on how touse the Punnett square (Y/N)?$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 parentsof known genotype%Rationale: The Punnett square lets us#determine the theoretical number of%offspring having different genotypes."Once we know the genotypes, we can%determine the offsprings' phenotypes. We can compare these theoreticalvalues to actual offspring. Procedure:1. Select trait.2. Select mates.3. Fill in Punnett square.4. Record the phenotypic ratio.5. Cross the peas.%6. Count offspring of each phenotype.$7. Compare the actual results to thepredicted results.$Purpose: To predict the frequency of&two traits in the offspring of parentsof known genotype$Purpose: To predict the frequency of three traits in the offspring ofparents of known genotype1. Select traits.HEIGHTTall: DominantShort: Recessive SEED COLORYellow: DominantGreen: RecessiveFLOWER POSITIONAxial: DominantTerminal: Recessive SEED FORMRound: DominantWrinkled: Recessive Tall: TT Tall: Tt Short: tt Axial: AA Axial: Aa Terminal: aa Yellow: YY Yellow: Yy Green: yy Round: RR Round: Rr Wrinkled: rrTall/axial: TTAATall/axial: TTAaTall/axial: TtAATall/axial: TtAaTall/terminal: TTaaTall/terminal: TtaaShort/axial: ttAAShort/axial: ttAaShort/terminal: ttaaTall/yellow: TTYYTall/yellow: TtYYTall/yellow: TTYyTall/yellow: TtYyTall/green: TTyyTall/green: TtyyShort/yellow: ttYYShort/yellow: ttYyShort/green: ttyyAxial/yellow: AAYYAxial/yellow: AaYYAxial/yellow: AAYyAxial/yellow: AaYyAxial/green: AAyyAxial/green: AayyTerminal/yellow: aaYYTerminal/yellow: aaYyTerminal/green: aayyTall/round: TTRRTall/round: TtRRTall/round: TTRrTall/round: TtRrTall/wrinkled: TTrrTall/wrinkled: TtrrShort/round: ttRRShort/round: ttRrShort/wrinkled: ttrrAxial/round: AARRAxial/round: AaRRAxial/round: AARrAxial/round: AaRrAxial/wrinkled: AArrAxial/wrinkled: AarrTerminal/round: aaRRTerminal/round: aaRrTerminal/wrinkled: aarrYellow/round: YYRRYellow/round: YyRRYellow/round: YYRrYellow/round: YyRrYellow/wrinkled: YYrrYellow/wrinkled: YyrrGreen/round: yyRRGreen/round: yyRrGreen/wrinkled: yyrr$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.$The highlighted boxes are not right."Do you want to correct them (Y/N)?"Reenter genotype and press RETURN."You must select a trait before youcan select mates. PhenotypeNumberTallShortAxialTerminalYellowGreenRoundWrinkled Tall/axial Tall/terminal Short/axialShort/terminal Tall/yellow Tall/green Short/yellow Short/green Axial/yellow Axial/greenTerminal/yellowTerminal/green Tall/round Tall/wrinkled Short/roundShort/wrinkled Axial/roundAxial/wrinkledTerminal/roundTerminal/wrinkled Yellow/roundYellow/wrinkled Green/roundGreen/wrinkled Select the phenotypic ratio thatreflects these results: 1. 100% tall 2. 100% short3. 1:14. 3:1 This ratio suggests which of thefollowing hypothesis:$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 recessivetrait.!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.$Look at this representation of human!chromosomes. Twenty-two pairs are"matched in both males and females.These are called AUTOSOMES.%The last pair, however, is matched in%females but not in males. These chro-&mosomes determine sex. They are called#SEX CHROMOSOMES. Females have two X!chromosomes. Males have one X andone Y.!Do you think that the sex chromo-#somes determine anything other than gender (Y/N)?$Actually, these chromosomes not only!Right. These chromosomes not only$determine sex but also carry alleles"for other traits. The Y chromosome&has fewer genes than the X chromosome.%Therefore, males have only one allele'for some traits while females have two.%Color blindness is an example of such&a SEX-LINKED TRAIT. It is carried only%on X chromosomes. Let's see how it is inherited.%A female who does not carry an allele"for color blindness has a genotype&X@WC@ZX@WC@Z. Her eggs are all X@WC@Z.See the Punnett square above. X@WC@ZX@WC@Z&Press @R to see the results of a cross#between a normal woman and a color- blind man.X@WC@ZX@Wc@ZY%chromosome and one Y chromosome. Only#the X chromosome carries a gene for"color blindness. Since this man is'color blind, his gametes are X@Wc@Z andY.Now fill in the Punnett square.Right.$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$dominant trait. But appearances can be deceiving.!Now let's see what happens when a&color-blind woman (X@Wc@ZX@Wc@Z) mateswith a normal man (X@WC@ZY).Here is the Punnett square.$Only the females have normal vision!Why are the sons color blind?$Color blindness is dominant in males%They inherit their one recessive gene for this trait from their mother%Right. You have discovered one of the%characteristics of sex-linked traits.$Sons inherit these traits from their%mothers. Normal vision is still domi- nant, but the sons have only oneallele -- a recessive one."No. Sons inherit sex-linked traits$from their mothers. Normal vision is&still dominant, but the sons have onlyone allele -- a recessive one.$Dominance is not universal, however.!Press @R to see what happens when you cross these two snapdragons.%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.$This cross shows that the genes must!be discrete, not blended: Red andwhite both reappear in the F@2 generation.!What explains the pink offspring?#Alleles for both red and white pig-ments are expressed.A third allele is expressed."No. Both red and white alleles for%Right. Both red and white alleles for#pigment color are expressed. If you%were to magnify the snapdragon petal, you would see both pigments dis-$tinctly. But to the eye, they appear to blend."Inheritance of pink color in snap-%dragons is an example of CODOMINANCE,!also called INCOMPLETE DOMINANCE.  " .$A color-blind man has sperm that are%either X@Wc@Z or Y (Y contains no al-leles for color vision).$This cross predicts that half of the"children will be males with normal$color vision. The other half will be"females who have normal vision but%all of whom carry an allele for color blindness.$But they carry one recessive allele.All males are color blind."This is a Punnett square showing a%cross between a homozygous red parent"and a homozygous white parent. Alloffspring are pink.RR@W'@ZRR@W'@ZR = allele for red petals R@W'@Z = allele for white petals#C = allele for normal vision (dom.)%c = allele for color blindness (rec.)both alleles are expressed."Codominance, incomplete dominance,#and sex-linkage are all examples of#the complex way in which traits are%passed from generation to generation. X@Wc@ZX@Wc@Z%According to Mendelian genetics, what%percent of the offspring of the above$parents should have black bodies and purple eyes? 25 percent 50 percent"Right. As the above Punnett squareNo. As the above Punnett square%predicts, 25 percent should be black-%bodied, purple-eyed flies. If you did#a TEST CROSS to verify this predic- tion, however, the results might surprise you.$Remember, in a test cross you mate a%dominant individual with a homozygous&recessive. Press @R to see the resultsof crossing these flies.#Practically half the offspring have$black bodies and purple eyes! What's going on?The flies are mutationsThe two genes are linked%Right. The two genes are linked. They"No. The two genes are linked. They%are found on the same chromosome. The%genes Mendel studied were all on dif-!ferent chromosomes. They assorted%independently. Not all genes are likethat.$In fruit flies, brown body color (B)$dominates over black (b) and red eye$color (R) dominates over purple (r).%The genes for black bodies and purple#eyes in fruit flies are on the same!chromosome. And usually traits on!the same chromosome are inherited together.#Now look at the fruit fly offspring&again. If black bodies and purple eyes$are always linked, shouldn't exactly%half the offspring have these traits?!Yes, you would think so. But they&don't. Something else is going on. You%can find out what is happening by re-examining meiosis.%Meiosis, if you remember, is the pro-#cess of cell division through which&animal gametes are formed. Press @R towatch part of meiosis proceed.%At prophase I, the chromosomes double#and then pair up. When they divide, the paired chromosomes sometimes#exchange parts and, hence, alleles.Press @R to see.%This is called CROSSING OVER. Alleles"are exchanged between members of a'homologous pair. Crossing over explains&the small number of black-bodied, red-"eyed and brown-bodied, purple-eyedfruit flies in the test cross.%Geneticists use crossing-over data to$determine where genes are on chromo-$somes. The closer two genes are, the&more likely they are to stay together.The further apart they are, the$greater the chance of crossing over.!Can scientists tell how close two%genes are on a chromosome by studyingcrossing over (Y/N)?!No. Geneticists use the number of!No. Only six percent of the fruit"Yes. Only six percent of the fruit&fly offspring were RECOMBINANTS, indi-"viduals unlike either parent. This!suggests that the genes are close$together and don't often cross over."Yes. Geneticists use the number of%recombinants to determine how far one#gene is from another. They can also%determine where on the chromosome the$genes are located. This picture of a chromosome is called a GENE MAP.&Would you like to learn more about how#geneticists create gene maps (Y/N)?%Gene maps have many uses. When scien-"tists know the exact location of a$gene, they can isolate and sometimes$reproduce it. They have done so with"genes for insulin and human growthhormone.&Knowing the exact sequence of DNA in a&chromosome also helps scientists study'viruses. Many viruses work by redirect-#ing the activity of the host cell's"DNA. Here, a virus attacks a cell.%But most fundamentally, a gene map is"a picture of DNA. The clearer thispicture, the better humankind's"understanding of the continuity of existence.&Geneticists divide the chromosome into'units. If two genes on the same chromo-'some are separated by crossing over one$percent of the time, geneticists say"they are one map unit apart on the chromosome.#six percent of the offspring in the"fruit fly test cross were recombi-#nants, how many units apart are thegenes for body and eye color?No. They are six units apart. That's right.$To figure this out, geneticists must"the genes are located, geneticists!consider three genes at once. The!question is: Which gene is in the middle? A THREE-POINT TEST CROSSprovides the answer."The results of a three-point cross show which genes cross over most%frequently and are therefore furthest$apart. Once you know which genes are"furthest apart, the remaining genemust be in the middle.&To do a three-point cross, geneticists%mate one trihybrid and one homozygous%recessive individual for three traits that are on the same chromosome.%If body and eye color are linked, and#But where on the chromosome are thegenes actually located?$Body color, wing type, and eye color#are on the same chromosome in fruit$flies. Press @R to see the amount of"cross over resulting from a three-point cross for these traits.TRAITS % CROSS OVER Body/wingWing/eyeBody/eye26206Which trait is in the middle? Body color Wing type Eye color%Right. The largest amount of crossing"No. The largest amount of crossing$over occurred between body color and&wing type. Therefore eye color must bein the middle.%The above gene map shows the relative!position, or LOCI, of these three%genes. To map the rest of the chromo-%some, geneticists would do additional$three-point crosses with these genes&and others with which they are linked."Using gene mapping and other tech-"niques, scientists have now mapped#most of the fruit fly's chromosomesand some human chromosomes. By doing a number of three-point&crosses, scientists can map the actual"positions of genes on chromosomes.bbppBbPpTrue or False:&If brown eyes dominate over blue eyes,#eventually everyone will have browneyes.TrueFalseRight."Actually, this statement is false.&As Mendelian genetics shows, recessive You can learn more about this bystudying what happens to genesin a POPULATION.$A population is a group of organisms$of the same species that live in the%same area. All the genes in a popula-$tion are its GENE POOL. The study of%what happens to these genes is calledPOPULATION GENETICS.$Scientists study population genetics!to predict how a population might%change over time. They use a mathema-%tical model called the HARDY-WEINBERGEQUILIBRIUM to help them.&Would you like to study the mathemati-#cal equation for the Hardy-WeinbergEquilibrium (Y/N)?%Many alleles govern eye color. All of!them must be recessive to produce'blue eyes. As Mendelian genetics shows,!recessive traits don't disappear.1. Mating must be random. 2. The population must be large.3. No migration takes place.4. No mutations take place.&5. Natural selection is not operating.$The Hardy-Weinberg Equilibrium holds$that the frequency of a given allele#in the gene pool will not change as%long as the above conditions are met.&Do you think that these conditions arealways met (Y/N)?$That's right. In nature these condi-&tions are not always met. As a result,#gene frequencies change and speciesevolve.&No. In nature these conditions are not always met. As a result, specieschange.#The gene pool for each of the above$animals has changed over time. Which!one would you like to learn aboutfirst?Type number and press RETURN.Press @R when done.%Select another animal to learn about:NONRANDOM MATING Remeber the fruit fly? Some male$fruit flies have white eyes. Females$don't choose them as often as mating partners. Laboratory studies show that the%white-eyed males in a fruit fly popu-"lation eventually become extinct.SMALL POPULATION%The cheetah was almost extinct at the&end of the last ice age. A small group"of survivors was isolated from the$larger population. The genes of such&a group may not reflect the proportion"of genes in the larger population.%This is called GENETIC DRIFT. Geneticdrift can lead to a new race or species. But it can also lead to&potential extinction. This is the casefor the cheetah."Unfortunately, cheetahs are almost%extinct again today. Today's cheetahs&are too much alike genetically. If one&is susceptible to disease, so are mostof the others. MIGRATION&The above table shows the distribution!of blood groups within a tribe of"American Indians that was isolatedfrom Europeans who emigrated toAmerica. Now look at how the distribution&changed within a group of Indians that"intermarried. This illustrates the%impact that migration can have on the gene pool.MUTATION#What do these birds have in common?#Their ancestors. Small mutations in#the beak structure of the Galapagos&finches lead to the evolution of seed- eating, cactus-eating, and wood-pecking finches, among others.$Charles Darwin studied the Galapagos$finches. They inspired his theory if evolution.NATURAL SELECTION&A short-necked giraffe would be an odd!sight today, but it wasn't at one#time. Today's giraffes evolved fromshorter ancestors. Some early giraffes probably had#slightly longer necks and legs than$most. They could reach leaves others%couldn't. Therefore, they were better"able to survive and reproduce. The$long-necked giraffe became the norm.$Do you think that the Hardy-WeinbergEquilibrium is most often:Maintained in natureViolated in nature%Right. The Hardy-Weinberg Equilibrium"No. The Hardy-Weinberg Equilibrium%is not generally maintained. However,$it is an important theory that helps!us understand what factors affect%allelic frequencies and how importantthese factors are.Mutations, migration, nonrandom&mating, natural selection, and genetic"drift all help explain how species evolve... Or change...Or die.According to the Hardy-WeinbergEquilibrium, the frequency of a%dominant allele plus the frequency of$the recessive allele is equal to allthe genes for the trait.%If two alleles govern a trait and the&frequency of the dominant allele is .2&(20 percent), the frequency of the re-'cessive allele must be .8 (80 percent).A(.2)a(.8)AA(.04)Aa(.16)&If the above frequencies hold for both male and female gametes, you can%predict the frequency of genes in the"next generation of the population.Press @R to see. the frequency of alleles has not%changed. You are observing the Hardy-&Weinberg equilibrium. This equilibrium is based on certain assumptions.aa(.64).04AA + .32Aa + .64aa = 1 .04AA + .16A .16a + .64aa.20A+.80a%= 1 \ FAH By adding the frequencies of re-!combined alleles you can see that"For example, consider the beaks on&these finches. Individual birds within$a population of finches have varyingdepths of beaks. 'This graph, called a BELL-SHAPED CURVE,#shows the range of beak depths that#occur in one population of finches. "The finches with smaller beaks can&crush only small seeds. If small seeds$become less abundant, the birds with!small beaks will die off. This isNATURAL SELECTION.#After several generations, the fre-"quency distribution shifts and the mean changes. Increased Decreased%Right. It has increased: More members"No. It has increased: More members"of the population have deep beaks.#The curve has shifted to the right.!This kind of natural selection iscalled DIRECTIONAL SELECTION."This change and others lead to the&emergence of one species from another.%It is one of the mechanisms that helpexplain EVOLUTION.&Not all traits are inherited following$Mendelian patterns. Some traits show#codominance; still others show sex-linkage."By carefully observing parents and$offspring you can determine the pat-tern of inheritance.Select the organism you want tostudy: DrosophilaHumans$PURPOSE: To determine the pattern of#inheritance of traits in Drosophila#METHOD: Select and cross Drosophila"mates, then count the offspring by$phenotype. Conduct a second cross if%necessary to determine the pattern of inheritance. ASSUMPTIONS:1. All offspring live. Each flyrepresents about 20 offspring."2. The number of offspring of each"phenotype represents the frequency$with which the phenotype would occurunder ideal conditions. PROCEDURE:1. Select a trait.2. Select a hypothesis.3. Select mates.!4. RUN the experiment to generate offspring.#5. Count the number of offspring ofeach phenotype and record yourresults.!6. Have the computer run a second#cross if necessary to determine thepattern of inheritance.7. Compare your data with your$prediction and draw your conclusion.METHOD: Select and cross twoRed/white eye colorRed/scarlet eye colorNormal/vestigial wing formBrown/yellow body colorBrown/plum body colorRed is dominant.White is recessive.Red is dominant.Scarlet is recessive.Normal wing form is dominant.!Vestigial wing form is recessive.Brown is dominant.Yellow is recessive.Brown is dominant.Plum is recessive.$Unless these alleles are codominant:Select a hypothesis to test:1. This is a sex-linked trait.2. This trait is an example of codominance.3. This trait is an example ofMendelian genetics.Red eyes White eyes Eye colorR=redW=whiteredwhite"How many female offspring have redeyes? $How many female offspring have whiteeyes? How many male offspring have redeyes? "How many male offspring have whiteeyes? Red eyes Scarlet eyes Eye colorR=red S=scarletredscarlet"How many female offspring have redeyes? How many female offspring havescarlet eyes? How many male offspring have redeyes? $How many male offspring have scarleteyes? Normal wingsVestigial wings Wing formN=normal V=vestigialnormal vestigial%How many female offspring have normalwings? How many female offspring havevestigial wings? #How many male offspring have normalwings? &How many male offspring have vestigialwings? Brown body colorYellow body color Body colorB=brownY=yellowbrownyellow$How many female offspring have brownbodies? %How many female offspring have yellowbodies? "How many male offspring have brownbodies? #How many male offspring have yellowbodies? Brown body colorPlum body color Body colorB=brownP=plumbrownplum$How many female offspring have brownbodies? #How many female offspring have plumbodies? "How many male offspring have brownbodies? !How many male offspring have plumbodies? !How many female offspring show anintermediate trait? How many male offspring show anintermediate trait? Type a number (0 - 8) and pressRETURN."The total number of offspring mustequal 16. Please reenter.Do you want to cross two of theoffspring (Y/N)?'Those mates will not provide sufficient$data to draw conclusions. Select one"dominant and one recessive parent.  YOUR HYPOTHESIS:This is a sex-linked trait.This trait is an example of codominance.This trait is an example ofMendelian genetics.$Do your data support your hypothesis(Y/N)?$That's right. Which of the followinghypotheses is correct?Good. You have successfullydistinguished sex linkage fromMendelian inheritance.#Good. You have successfully distin-"guished codominance from Mendelian inheritance.$That's right. This trait follows the!pattern of Mendelian inheritance.!No. This trait shows sex linkage.!No. This trait shows codominance. %No. This trait follows the pattern ofMendelian inheritance."PURPOSE: To predict the pattern of inheritance of some human traits#RATIONALE: If they know the pattern"of inheritance of a trait, geneti-#cists can predict the likelihood of!that trait being passed on to the#offspring of parents of known geno-"type. Such information is used forgenetic counseling.#1. You can't really do experimental"crosses with humans. This lab lets!you simulate what would happen if you could.%3. All individuals in the P@1 genera-tion are homozygous.Normal/hemophiliaNormal/sickle cell diseaseNormal/albinismNormal/Tay-Sachs disease $Normal clotting protein is dominant.Hemophilia is recessive.Normal hemoglobin is dominant.!Sickle cell disease is recessive. Normal pigmentation is dominant.Albinism is recessive.Normal is dominant.Tay-Sachs disease is recessive.Normal blood clotting Hemophilia HemophiliaN=normal H=diseasenormaldisease"How many females have normal blood clotting? "How many females have hemophilia?  How many males have normal blood clotting? How many males have hemophilia?  Normal hemoglobinSickle cell disease Sickle cellN=normal S=diseasenormaldiseaseHow many females have normal hemoglobin? !How many females have sickle cell disease? How many males have normal hemoglobin? How many males have sickle cell disease? Normal pigmentationAlbinismAlbinismN=normal A=diseasenormaldiseaseHow many females have normalpigmentation? How many females are albino?  How many males have normalpigmentation? How many males are albino?  NormalTay-Sachs disease Tay-SachsN=normal T=diseasenormaldiseaseHow many females are normal?  How many females have Tay-Sachs disease? How many males are normal?  How many males have Tay-Sachs disease? #When genes governing two traits are%located on different chromosomes they#assort independently. When they are"on the same chromosome they do notassort independently. They are linked. During meiosis, however,linked genes sometimes separate%through crossing over. The result is$a small number of recombinants, off-!spring unpredictable by Mendelian genetics.#You can use recombinant data to de-!termine whether genes are linked,#whether crossing over has occurred,"and how far apart two linked genesare on the chromosome.PURPOSE: To distinguish betweenlinked and unlinked genes in Drosophila"METHOD: Conduct a test cross, thenanalyze the offspring data to"determine whether or not the genes are linked. PROCEDURE:1. Select traits.2. Select a hypothesis.3. RUN a test cross.4. Analyze offspring data.5. Draw conclusions about your hypothesis.Pink eyes/striped bodySepia eyes/dumpy wingsYellow body/dumpy wingsEbony body/scarlet eyesPurple eyes/vestigial wingsPink eyes and striped body are recessive.%Red eyes and brown body are dominant.Dumpy wings and sepia eyes are recessive.Normal wings and red eyes are dominant.Yellow body and dumpy wings are recessive.Brown body and normal wings are dominant.Ebony body and scarlet eyes are recessive.%Brown body and red eyes are dominant.#Purple eyes and vestigial wings are recessive.Red eyes and normal wings are dominant.Select a hypothesis to test:1. These genes are linked.2. These genes are on separate%chromosomes and assort independently.$When you press @R, the computer will'generate the results of the test cross."This test cross uses one heterozy-"gous and one recessive individual.Press @R to begin. PHENOTYPENUMBER OF OFFSPRINGTOTAL OFFSPRING:100200Red eyes/brown bodyRed eyes/striped bodyPink eyes/brown bodyPink eyes/striped body437743Normal wings/red eyesNormal wings/sepia eyesDumpy wings/red eyesDumpy wings/sepia eyes23252725Brown body/normal wingsYellow body/normal wingsBrown body/dumpy wingsYellow body/dumpy wings22272328Brown body/red eyesBrown body/scarlet eyesEbony body/red eyesEbony body/scarlet eyes86141486Red eyes/normal wingsRed eyes/vestigial wingsPurple eyes/normal wingsPurple eyes/vestigial wings446644YOUR HYPOTHESIS:These genes are linked.These genes are on separate%chromosomes and assort independently.%Do these data support your hypothesis(Y/N)?!That's right. These genes are notlinked.#That's right. What is the number of recombinants?#What percent of the total number ofoffspring do they represent?Right.No. Try again.%Fourteen percent of the offspring are recombinants.#Twelve percent of the offspring are recombinants. How many map units apart are thegenes?Very good. You have correctly$interpreted the results of your two-point test cross.No. Try again.%No. Fourteen percent of the offspring were recombinants, therefore thegenes are 14 map units apart.#No. Twelve percent of the offspring&were recombinants, therefore the genesare 12 map units apart. PROCEDURETHEORY$Wrong. The genes are linked. What isthe number of recombinants?No. There are 12 recombinants.No. There are 14 recombinants.No. There are 28 recombinants. Red/pink eyesBrwn/striped bodyNorm/dmpy wingsRed/sepia eyesBrwn/yellow bodyBrwn/ebony bodyRed/scrlet eyesNorm/vest wingsRed/purple eyesNo. These genes are on separatechromosones. They assortindependently. TRAITS: Lets you select a trait.!MATES: Lets you select two mates. RUN: Lets you carry out a cross.EXIT: Returns to main menu.it.!MATES: Lets you select two mates. RUN: Lets you carry out a cross.EXIT: Returns to main menu.1. Select a wavelength.2. Run the experiment. Read themeter.3. Record your results.$4. Repeat the experiment with a dif-!ferent wavelength. Complete seven%trials for good experimental results.Color WavelengthViolet 400 - 455 nmBlue 456 - 495 nmGreen 496 - 560 nmYellow 561 - 575 nmOrange 576 - 595 nmRed 596 - 700 nm400410420430440450460570480490500510520530540550560570580590600610620630640650660670680690700 Wavelength:Color:&When you press @R, the experiment will&begin. Read the meter to measure light absorbed.#You must select a wavelength beforeyou can run the experiment.!The range of possible readings is 0.0 to 2.0.%Right. Do your data support either ofthe other hypotheses?$1. Green light is absorbed and might!therefore trigger photosynthesis.%2. Extracted pigments absorb light ofall wavelengths.#3. My data do not support either ofthe above hypotheses.Check the data again. Plantsdo absorb red light.do not absorb green light.do not absorb all wavelengths."Do your data support either of theother hypotheses?Yes. Plants absorb red and blue$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 ' !"#$%&'()*+,$K11 1}1 11 1KNHMO HMO-HGenetics: Disk 2G" HMdP-H;<Other PatternsGW of InheritanceGxhiWW>48gA< Crossing OverGxh iWW>4 8BA<Hardy-WeinbergGW EquilibriumGxhiWW>4u8 AjkcdK HMOxHO*H*QH)Q9ZHBbRrGZHbbrrGa2CEWEW EefaMFO 1.H MOHD Fa2CEWEWEWE1 I;<EW> 18A<E> 18AB 1 HMlOaMFO(sBRpqrvtwF HFemaleG-HMaleG-HM$=Pa2C9E8EW EW EW EW EW Eefod HMlOaMFO*H*QH)Q9ZHBbRrGZHbbrrGa2CEWEWEWEUUd HMlOaMFO.H MOHD F<HTRAITSG2 IPERCENTG< HMO 1 J Black/purpleGn I42GW<I Brown/redGn I42GW<I Black/redGn I8 GW<I Brown/purpleGn I8 GW 1a2CEWEWE 11 I;<EW J> 18A<E> 18AB 1@aMFOa2C9E8EWEWEWEWEWEefud HMlOaFOF2H#QP(HB G IR GPPHB G IR Ga3CEWEWEWEWEefd HMlOaMFO<HTRAITSG2 IPERCENTG< HMO 1 J Black/purpleGn I42GW<I Brown/redGn I42GW<I Black/redGn I8 GW<I Brown/purpleGn I8 GW 1a3CEWEWEW Eef'daMFOa3C EW EW EW Eefkd HMlOaMFOZH&QHEarlyGH Prophase IGa3CEWEWEWEU\aMFOZH'Q\HMFOHMid-GH Prophase IGZH(QefdaMFOa4CEWEWEWEWEUZH$Q>HMP\aMFO HMlOhH1Q \2Q \3Qef|daMFOa4CEWEWEW EW EW EeaMFO HMlOHMid-GH Prophase IGZH(Qfd HMlOaFOFH%Q2HMO2H Glued eyesGZHM-O<H Hairy bodyG2HMO2HJavelin bristlesGa4C EW EWEWEWEWEe HMlOaFOhH1Q \2Q \3Qfd HMlOaMFO<HTRAITSG2 IPERCENTG< HMO 1 J Black/purpleGn I42GW<I Brown/redGn I42GW<I Black/redGn I8 GW<I Brown/purpleGn I8 GW 1a4CEWEWE I[daMFO HMlOFH%Q2HMO2H Body colorGFHM-O(H Eye colorGZHM OZH OdHOdHDGENE MAPGa5C9E84FWEWEWEWEWEefFH%Q2HMO2H Body colorGFHM-O(H Eye colorGZHM OZH OdHOdHDGENE MAPGaMFOa5CEWE[93n d HMlOaFOFH%Q~H/QAHMOa5C EW EW EW EW EWEe HMlOaFOfPd HMlOaFOTH&Q8'H.Qa5CEWEWEWEWEefd HMlOaMFOPH%QNSHM PSHMON?H,Q;HM1Pa5CEWEWEWEWEef5d HMlOaFOFHMOFHONHOJJ IO98GJ#IMO1IO#8HO!H One map unitG HRO*I Oa6CEWEWEWEWEWEefd 1 HMlOaMFO(H MO`ID FFHMOFHONHOJJ IO98GJa6CEWEWEW EW EdY IZ*aMFOGJ#IMOIMO#8HdO(!H Six map unitsG( HTOMI Oa6C*9 E8 EWEWEefd HMlOaFOFH%Q2HMO2H1 GZHM-OYH2 G2HMO2H3 Ga6C EWEWEWEWEef1daMFOa6CEWEWEWEWEWEefdaMFOa7C!EW"EW#Ee HMlOaFOFH%Q2HMO2H1 GZHM-OYH2 G2HMO2H3 Gfd HMlOaFO*H+QH)Q.ZHBbRrWwGZHbbrrwwGa6CEWEWEWEe HMlOaFOFH%Q2HMO2H1 GZHM-OYH2 G2HMO2H3 Gf7daMFOa7CEWEWEWEWEU HMlO7C2(HE2 IE2 HMO 1 JE I EW2I E I EW2I E I EW 1aMFO.H MOHD Fa7CEW1( I;<EW> 18%A<EW> 18A<E> 18ABaMFON28HE I EWNa7C9E8EWEWEWEe HMlOaFO*H+QH)Q.ZHBbRrWwGZHbbrrwwGfd HMlOaMFO2HMO(2HO2HO:HOJJCHMO)CHOGHO,CHOCHO,GHOH Body colorG0HO*0HO*H Eye colorGH Wing typeG0HOIH6 GIH20GUHMOUHOUHOxWH26GaMFOa7CEWEWEWEWEWEe HMlOaFO*H+QH)Q.ZHBbRrWwGZHbbrrwwGfd HMlOaMFO2HMO(2HO2HO:HOJJCHMO)CHOGHO,CHOCHO,GHOH Body colorG0HO*0HO*H Eye colorGH Wing typeG0HOIH6 GIH20GUHMOUHOUHOxWH26Ga7CEWEWEW Eef3 cdK HMOxHO.H MOHD F HQ 1a8CEWEWEWE1 I;<EW> 18A<E> 18AB 1FaMFO H!Qa8C9E8EWEWEWEWEef!aMFOa8C EW EW EefYd HMlOaMFO* HQ HQa8C EWEWEWEWEWEefBd* HQ HQaMFOa8CEWEWEWEWEef(aMFOa8CEWEWE[93&d HMlOaMFO9C1HEWEWEWEWE 1aEWEWEW Ee HMlOaMFOf-aMFOa9C EW E[9181aMFOa9C9EWEWE8 EW EWEWEe HMlOaMFO9C1HEWEWEWEWE 1f^d HMlOaMFO*HQiIQIQ<C 1aEWEWEWEWEefhd HMlOaMFOFH8QH FrequencyGdHDDepthG<CaEWEWEW Eef,daMFO<Ca EW EW EW EWEef{daMFOFH9Q<CaEWEWEWe HMlOaMFOFH8QH FrequencyGdHDDepthGfodaMFO<CaEWEWEe HMlOaMFOFH8QH FrequencyGdHDDepthGfd HMlOaMFO*HQiIQIQ<CaEWEWEWEe HMlOaMFOFH8QFH9QH FrequencyGdHDDepthGf5Rd HMlOaFO(HD A + a = 1Ga;CEWEWEWEWEe3fZdaMFOg2HF2HF@HD .2 + .8 = 1Ga;CEWEWEW Eefd HMlOaMFOFHMFONI-HOg!HAOo#H;C E I EI2H EWW ENaEWEWEWEWEUNo2H E I EWWoI E IENe HMlOaMFO(HD A + a = 1Gfd HMlOaMFOHD;F2FH;FFH;FDdH;F6UH0QdH;FUH0QdH;FdH;Fa;CEW EWEWEWEWEef3QKNHMO HMO-HGenetics: Disk 2G" HMPP-H;<Lab: Determining GW Patterns of InheritanceGhiWW>4~B8\A<Other Patterns of GW InheritanceGhiWW>4f!8"A?5AjKNHMO HMO-HGenetics: Disk 2G" HMKP-H;<Lab: Gene LinkageGh iWW>4w28xA<Review of CrossingGW Over and LinkageGhiWW>4G89A?5AjOcdK HMOxHO*HQHQa1CEWEWEefWdaMFO HMlOHQH AutosomesGa1CEWEWEWEef[daMFOUHQUH M$P 1a1CEW EW EW EW EW E 1ef^d HMlOaFOTH&QHQU2HM2Oa1CEWEWE I[KK H4QF HM4iP H4iP HMOyHO DHFEMALEG  HODHMALEG H Oa1C9E8EWEWEWEWEWEefd HMlOHFOxHMOHQHQ#_H1FW1F;HC GHc Ga1CEWEWEWEefd HMlOaFOFHMx8OdHM6OH.HtO1CNl#H$E#H$EN 1a1CEWEWEWEefFd HMlOaFOH5QH6Qa1CEWEWEefd HMlOaMFOFHMx8OdHM6OH.HtO1CNl#H$E#H$EN5H%ENGH&ENa1C!EW"EW#EUNl5H6E5H6ElGH7EGH7ENefd HMlOaFO1C&H MOFH4E I5EF HO1U0H6EW7E 1a9EW I:EW;EW<EW=E`ID>E"11+Y 1HZ,,+94&+1+>983J(aMFO1CHM Oa?ENH MxO`ID FUaMFO&H MOa9EW I:EW;EW<EW=E`ID>EYHZ,,+9aMFOHM O+9H3 Ga@EWAE8H1 GaBEWCENH MxO`ID FU>9LaMFO&H MOa9EW I:EW;EW<EW=E`ID>E5]daMFO0H Normal femaleG>H Normal maleGa1CEWEWEWEWEWEefaMFOaFEWGE[aMFO9HEWIEWJE8KEWLEWMEWNEWOEWPEe HMlOaFO1C MFH4E I5EF HO1U0H6EW7E 1f^d 1 HMjOaFOFHQNUH1FN HQNH17FNCHQNNH????GNTCHQNcNH????GN<HQNMH????GN<HQNMH????GNa1CQEWREWSEe HMlOaFO1C MFH4E I5EF HO1U0H6EW7E 1fd HMjOaMFOFHMx8OdHM6OH.HtO1CNl#H%E#H%EN5H$ENGH&El5H6E5H6ElGH8EGH8ENa1CUEWEWEefd HMlOaMFO.H MOHD F*HQHQa1CVEW1 I;<WEW J> 18A<XEWYE> 18ABZ HMjOaFOFHQNUH1FN HQNH17FNCHQNNH16FNMCHQN\MH16FN<HQNMH18FN<HQNMH18FNFHM84P?H84P?H83Pa1C9`EWaEWbEWcE8ZEW[EW\EW]EW^EW_Eefd HMjOaFO1C8HQH Q8HF((HE(HE-ZHEWEadEWeEWfEU-ZHMOodaMFOCH QTCH QCH QCH QFHFa1CgEWhEWiEWjEWkEWlEUd HMjOaFO8 H Q H Q8HF\CHQTCH QCH QCH QFHFa1CmEWnEWoEWpEe HMjOaFO8 HQ H Q8HFfxdaMFO.H MOHD Fa1CqEW1 I;<rEWsEW J> 18A<tE> 18AB HMjOaFO8#H QH Qd4HM>Oa1C9vE8uEWwEWxEWyEWzEW{Ee HMjOaFO8 H Q H Q8HFCHQTCH QCH QCH QFHFf#daMFOa1C|EW}EW~EWef!d HMjOaMFOFHMx8OdHM6OH.HtO1CNl#HE#HEN5HENGHEl5HE5HElGHEGHEN(_HEWEa1CEWEWEWEe HMjOaFO8#H QH Qd4HM>OfVd HMjOaFO2HQgHQ2HQa1CEWEWEWEef5ucdK HMOHDOCLAB: GENE LINKAGEGWW IEWEWEWEWEWEWEWEW EW EW Eef9dK HMO HOC EW EWEWEWEefvdK HMO HOCEWEWE  HM?O. H MO  HEWEWEWEefjdK 1 HMO HOCEWWEW JEW JEW JEW JEWE 1efK1!1"4?NaMdOH O HMO-H"Cg;<TRAITSG h>448A I<RUNG h>4978A I< ? G h>4nA8_A I<EXITG h>3@8RCTH MoOaEWEWEWE`IDEUaM FO #HMPO #HSFI NO. OFFSPRINGG  H MKOIUO  H 1S CEIEW IEIEW IEIEW IEIEW I JSFI 9 SF8SF 11!1"A 1aYCEW J 9E8 EWEW JEWE[aM FOaYC  /9V 9&9 EWE8[CEWEWE8$.9 EWE8[CEWEWE8p&H MO`ID Fa 9!9  EW E8 [CEWE8.9  EW E8 [CEWE3Y:0TH MoO`ID FU4@4?5dY IZ))YCaM FOa )0 )0/ ) 0/9J>HMP&H MO`ID Fa EW EW E3 <981ELH MO`ID FUaM FO&H MO`ID FaHow many of the offspringGWare recombinants?G3Y:r>HMP&H MO`ID Fa[C 9E 9E 9EWYC EW E IdYZ( \aM FOLH MO`ID FaYC (0 (0/ ( 0/9 E8   /9E8EWEU HMfOa FOHMVPHPERCENT RECOMBINANTS: G 912G814GFHMOFHONHOJJ IO98&H MO`ID FaZCEWEY IZ*waM FOLH MO`ID F  /9GJ#IMOIMO#8HO<!HFourteen map unitsG< H|OzI OaZC*9EWEWE8EWEW E8GJ#IMOIMO#8HO4!HTwelve map unitsG4 HvOpI OaZC* 9EWEWE8 EW EW EU4@4?5KN HMOHOHVP"*9HTRAITS: G[C 9E I E 9 E I E 9 E I E 9 E IE 9E IE"*9 #HSFI NO. OFFSPRINGG  H MKOIUO  H 1S CEIEW IEIEW IEIEW IEIEW I JSFI 9 SF8SF 1511 1"1 1)1(5"!09ta You have not Gcompleted the entireGWlab sequence. Are you sureG that youGWwant to exit (Y/N)? G[.9344@5{ HMT#O HMPP;  H< ProcedureGLh> 4A3?8'A H<TheoryGLh> 4G3?8AjK 1 HMOFH MO`ID" F HOCEWWEW JEW JEW JEW JEWEU 15cdK HMO 111' H=CLAB: DETERMINING PATTERNS OF G INHERITANCEGW JEWEWEWEWEWEWEWWEW EW J;(I< E H< E ?A(H MO HD FB 1!dK HMO! > C HEWE!9 VMO. H MO  HE!9 VMOWEWEWEWEW!9 EW EWefdK HMO H>CEWW!9< EW EW J EW EW EWEW JICEWE8KIC EW EW EW EW J>C EW EW EWEW JICEWEefdK 1 HMO H>CE  HE  HE  HE  HEWE  HEWEWE  HEWEWE  HEWEW 1ef K1"4FNaMdOH O HMOHg;<TRAITSG h>4H8A I<MATESG h>4J8A I<RUNG h>4N8aA I<? G h>4T8@A I<EXITG h>3T8AjKN HMOHOX HjP"*9_! @  CMHEM VOYHEcHEmH I=Intermed.G"*9(HMatesGM VO! @  C4HF: GI&$0$/9E$&09  Intermed.G$9E>HM: GI&#0#/9E#&09  Intermed.G#9E"*94EN5-HMThOZHMhO  HMO HM P H P# H MOHD"F HD"F! ? C1   H!1HM O ET 98L 9 1  91 8& 9  1  &91 888!9v 91&RWp 91&RSp 91&NVp 91&BYp 91&BPp8\ 91&NHp 91&NSp 91&NAp 91&NTp&HM O HO H@CE;( H<E>1%A H<EWE>1%A H<EWE>1%A,H MOHD FB1"3FW"9NaMdO'CEWEJH MOHD FUaMdO54HMTOZHMhOF HMOL HM PM H PaM FOaYou will select the female Gfirst.GJH MOHD FUaM FO(HMatesGM VOa! ? CaMdO?FW J  EW  E! @  C4M1$4HF: GI$9E$9E4M1#>HM: GI#9E#9E$#9qaMdOGCEWEWEWEWEJH MOHD FUaMdO4HMTO8F HMOX HMjPaMdO1"51#H MOHD"FHD"FRHM OET :3  /9 1&918883RHM O#HMO915U$q#rZHMhO1b " H 1:5`"9TaMdO'CEWEWEWJH MOHD FUaMdO54ENOaMdOH M O HFF GFF! @  C1 11 aMO9A  E! / /9 W  E  1 8%F FWF FYZ  1  &: 1:SaMdO 98FCEWEJH MOHD FU8@! @  CKM HOHOH YOUR RESULTS:GH9P)HO7HOEHOH9OH9O$HEHDFemaleGHDMaleG$,HE$:HE$HH intermed.G1, HD G , HD G 1:HC HE( H%9 EWE%9 EWE%9E"97 HGFWGF[1"91#1$4F3NHC HM(OEWE[&%:.9 1 HM ZO EW E( H;%9 H<E>1%A%9 H<EWE>1%A%9 H<EWE>1%A,H MOHD F1B 11HC HM ZO%&9{.92No. Your original hypothesisG was correct.G8@%9 EWE%9EWEWE%9 EW EW E81&9 EWE&9 EWE&9EJH MOHD FU1"1'3F~'9ta You have not Gcompleted the entireGWlab sequence. Are you sureG that youGWwant to exit (Y/N)? G[.93E5 HMT#O HMP!P;  H<CommandsGLh>44U3FA H< ProcedureGLh>4U3FA H<TheoryGLh>4f!3FAjqK HMOFH MO`ID" F H\CEW JEW JEW JEWU5K 1 HMOFH MO`ID" F  H>CE  HE  HE  HE  HEWE  HEWEWE  HEWEWE  HEWEWU 15U 1515H+G \ HADPG \HATPGgh5efK HMO.H MO2HLAB: ABSORPTION SPECTRUMGWW I?CEWEWEWEWEW EW J12I;< EW J>3yW8A< E>47i8A`ID FBfK1( 1 HMO H?C EW EWE. H MO  HEWEWEWEWWould you like to see the GW set-up (Y/N)?G[94 d 1CfK HMO H?CEWEWEWEWEWEWEghfK HMO.H MO H-CEWE H1@C;<EWEWW> 1"8DA<EWEWEWW> 1"8'A<EWE> 1"8AHD FBbfK HMO H-C#EWWACEW JEWEW JEW JEWEWEghK4i1){r4eNa JM <OH O HMO H.Ci;<E j>4sZ8A I<E j>4\8A I<E j>4w]8aA I<E j>41e8AA I<E j>3`8!AlHM<Oa JM <OHM-OHM-P )09CHH MO`ID0Fa J@CEWEWEUa JM <O.H MO`ID FNHnmGNHAQ H400-G6H456-G`H496-GH561-GH576-GH596-G H 455G6H 495G`H 560GH 575GH 595GH 700GNAC1 HM O   1  G -HQT 9 ~8R 9 1  91 8' 9  1  &91 888oN1NHM<O5HM<O;Hm HMrONH YOUR RESULTSGN(HMUOnoqpFH MO`ID0FUN&91&54e5tHM<O 9Ya J0C EW EWE[94`a JM <Oa JEWE[.93]ar4i4e5H9?HH MOACa J:EW;E`ID0FU5RH MwO HD0FWDPress RETURN to cancel.Ga JAC7EW8EW9ET: 958HM<OyH ^OM HN2HM"O@2H"ONNN2HM"O@2H"ON  1 a JM <O'H MO`ID0CEa JEWENxN9Wa JM <O9H MO`ID1Fa JAC<EW=EU8V8NwHM<O5 &9,4ia J0C EW EWE[94`3]a8e 9-a J1CEWEWE[94ir3N3Y8/a J1C EW EW EWE[94ir3N3Y1&a JM <OHH MO`ID0F HMrONH IDEAL RESULTSGN(HMUO-HNM<O;HmnoNUa JM <O3\Kf HMOdHOA HMUO_!Hmnoqp2H MO`ID F 1aJ@CYOUR HYPOTHESIS:GW J"9  EW EW"9 EW EW EW"9EWEWWE H2CE[ 1aNaJM ZOaJBC99"9EWE1$gh8 EW"9 EW8 EWgh$94ir3NN"9BC EW EW1$gh8K-C HMO}HO+H MO`ID F H1.G@C H"9 EWE8  EW EWWI2.GI EW EW EWWIBCEWIEa J2CEWEWYZK$98?a JM ZOa JBC9 EWE8 EWEgh4ir3N$K HMOHOHH MO`ID0F HQM H!QNpHD2.0Gp.HD1.5GpGHD1.0Gp[HD0.5GpjHD0.0GN-H QZHLIGHTGxH(9 EXTRACTG8 PIGMENTGZxHMETERG7HMO+HPOU5(9  H8 HMT#O HMP!P;  H<CommandsGLj>4gA H< ProcedureGLj>(94u84hA H<TheoryGLj>4CAlKNM HOHO HrO(HMUO7HMOM H!QNpHD2.0Gp.HD1.5GpGHD1.0Gp[HD0.5GpjHD0.0GN -HQ@7HMO+HPON,HnmG,H AbsorbanceG:H<vyzN9RHM-OHM-PNHnmG8CHM O GN9wHMUOHMUP H9Q9CQ9BQH9NC# F8  HQ5KNM HOFH MO`ID0FH3CBFW J(9DFW JEWEW EW J EW EW J EW EWU5K HMOFH MO`ID0F H-C#EWWACEW JEWEW JEW JEWEWEU5111 1 1 5fKr1( 1 HMO HCCEWEWE?C. H MO  HEWEWEWEWWould you like to see the GW set-up (Y/N)?G[94 d 1>fK HMO HCCEWEWEWEWEW EghfK HMO.H MO H-CEWE H1CC;< EW EWW> 1"8DA< EWEWW> 1"8,A<EWEWE> 1"8AHD FBfK HMO H-C#EWWCCEW JEW JEWEW JEW JEWEW EW J!EW"EghK4i1){r4eNa JM <OH O HMO H.Ci;<E j>4sZ8A I<C#F j>4ml8~A I<E j>4\8^A I<E j>41o8>A I<E j>41e8A I<E j>3~p8Ala JM <O) 09Ra JYou have completed this lab.GHH MOID0FU3~p  /98?a JCC(EW)EW*EW+EW,E[.951 11s4eHM<O 91 11a JWould you like to compare G your resultsGWwith the ideal before you G begin theGWnext series (Y/N)?G[94:*N1 1 ) 1)t4ea JM <OHMUOHMUP.H MO`ID FNCC$1HMP OE H$%/9Q&9CQ'9BQT 9 #8R 91$9'18' 9 1'&9$1888QNNHM<O5HM<O 9a J0C EW EWE[94`)93-qa JM <OHH MO`ID0Fa JYou are now ready to do G your next GWseries of trials. GUt4e1) 1)1 11 1 5W/9DHH MOCCa J-EW.EW/E`ID0FU537^ )09:1 1)11a J0C EW EWE[94`3-q 9)a J1CEWEWE[94i3N3fka J1C EW EW EWE[9 4i3N83fkKf HMObdINCEN IE( HMOJH MO`ID0FHThe next screen will display G graphsGWfor all of your trials. 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