Solution Manual for Method and Practice in Biological Anthropology: A Workbook and Laboratory Manual for Introductory Courses, 2nd Edition

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Chapter 1: The Scientific MethodReadingQuestions1.The idea that the universe is controlled by a supernatural force or deity is best described asa(n): (a) empiricism, (b) cause and effect,(c) teleology,(d) scientific theory.2.The first step in the scientific method is: (a) formulation of a hypothesis,(b) observation ofan event, (c) setting up an experiment, (d) theorizing on the likely result.3.Scientific reports of experiments are usually reported by the investigators in: (a) newspapers,(b) textbooks,(c) scientific journals, (d) magazines.4.The basic assumption in science that all humans experience events in the same way throughtheir senses is called: (a) uniformity in space and time, (b) natural causality, (c) cause andeffect,(d) common perception.5.A scientific statement that is based on experimental data and has some validity is known asa(n):(a) conclusion,(b) theory,(c) hypothesis, (d) explanation.6.The condition or event that may change in an experiment is the: (a) independent variable, (b)controlled variable, (c) original observation,(d) dependent variable.7.True orFalse: The results of an experiment do not have to be repeatable.8.Trueor False: An experiment wherein the researcher cannot control all of the variables,common in animal behavior studies, is a natural experiment.9.True orFalse: The variable that researchers try to keep the same for the experimental andcontrol groups is the dependent variable.10.True orFalse: Evolution is a popular hypothesis in biology, which needs further support todemonstrate its validity.

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In-Class ExercisesExercise 1Some people claim that epileptic seizures are the result of a supernatural force being directed at aperson for punishment of past behavior.Is this a statement of cause and effect or teleology?Does it violateany ofthe assumptions above?If so, which one?Teleology,Yes,it violatesnatural causality.Others claim that epilepsy is the result of neurons misfiring in the brain of afflicted individuals.Does this statement represent cause and effect or teleology?Does it violateany of theassumptions above?If so, which one?Cause and effect,Noit does not violate any assumptions.Exercise 2An experiment is done to test the effect of a new experimental drug for high cholesterol.A group of 200volunteers are separated into two groups of 100 each. Both groups are instructedto followa similar diet and activity level. Group 1 is given the experimental drug daily for 90days, while Group 2 is given a placebo. The individuals in the groups do not know whether theyare taking the new drug or the placebo. All participants are tested at the start of the study for theirserum cholesterol levels. The average for Group 1 is 310mg/dland the average for Group 2 is302 mg/dl.After 90 days,all participants’serum cholesterol is tested with a blood test. The averageserum cholesterol level for Group 1 is 299mg/dl and the average for Group 2 is 300mg/dl.Using this information answer the following:a.What is the hypothesis being tested?Theexperimental drug will lower the cholesterol ofindividuals who have high cholesterol.(Hypotheses may be written differently,using anH0and HAscenario)b.What is the dependent variable?Theserumcholesterollevelc.What are the independent variables?Thedrugand placebod.Which variables are controlled?Group size,diet and activities

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e.Which is the control group?Group 2(given the placebo)f.Did the experiment produce data that supports the hypothesis?Not really. The meanvalues for Group 1 did decrease more than forthe control group (Group 2), but not bymuch (cholesterol is still quite high).Exercise 3An experiment is done to test the effect of a high-fat diet on mice.In all, 50weanlingmice are separated at random into two groupsof 25 each.At the start of the experiment all miceweight approximately the same amount, about 20 g.Group 1 is fed a normal diet with balancedamounts of protein, carbohydrates, vitamin supplements and fat.Group 2 is fed the same amountof protein, carbohydrates, and vitamin supplements, but given a much higher fat content.Thecages are cleaned and mice are given fresh food and water daily.After 6 months all mice are weighed.The average weight in grams forgroup 1 was 8.2g.The average weight for group 2 was 12.6g.Using this information answer the following:a.What is the hypothesis being tested?A diet high in fat will cause mice to gain more waitthan one with a normal, balanced diet.(Hypotheses may be written differently, using anH0and HAscenario)b.What is the dependent variable?The weight of the micec.What are the independent variables?The amount of fat in the dietd.Whichvariables are controlled?The amount of protein, carbohydrates, vitaminsupplements and fat, how often the cages are cleaned, and how often the mice are givenfresh food and water.e.Which is the control group?Group 1(lower fat diet)f.Did the experiment produce data that supports the hypothesis?Yes, the average weightfor Group 1 is much less than for Group 2. The Group 2 mice are heavier.Exercise4An experiment is done to test the effect ofartificial lighton geraniums.Seventy-fivegeranium seedlings are grown in alaboratory.The plants are separated into five groups of 15plants each. Thefollowingtable shows the groups, how much light each receives per day, and

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the average height of the plants after 180 days in thelaboratory.All plants are fed the sameamount of water and fertilizer daily.Group 4receives as much light as all the other plants in thelaboratory, which are not part of the experiment. Thus, 16 hours is average. Groups 1 and 3receive less light than average, and Group 5 received more light than average.Group #Hours of lightper dayHeight at 180 days135.0 cm2614.5 cm31229.2 cm41636.1 cm52425.4 cmUsing this information, answer the following:a.What is the hypothesis being tested?The more light a geranium gets per day, thetalleritwill grow. (Hypotheses may be written differently, with an H0and HAscenario).b.What is the dependent variable?The height of the geraniumsc.What are the independent variables?The amount of light per dayd.Whichvariables are controlled?The amount of water and fertilizer(also group size,length of study)e.Which is the control group?Group 4(with the average light per day)f.Did the experiment produce data that supports the hypothesis?Yes, up to 16 hours oflight per day (the more light geraniums received, the taller they grew). But, whengiven 24 hours of light per day, growth was reduced.Exercise 5Aseries of observations that might be made by a biological anthropologistare listed atthe end of this paragraph. Working in teams, choose one observation from the list, formulate avalid, testablehypothesis,and roughly design an experiment to test your hypothesis.In yourwork, state your hypothesis, dependent variable, independent variable, and control variable(s).*The answers to this question can very greatly. The hypotheses listed are just examples.*a.Children from low income households show evidence of malnutrition.

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H0:Low income families cannot afford to purchase enough fruits and vegetables so theirchildren are malnourished.b.In most humans, the right humerus (upper arm bone) is larger than the left humerus.H0:The right humerus is larger in most humans due to most humans being right handed.c.Expectant mothers who smoke often have low-birth-weight babies.H0:The chemicals in cigarette smoke willslowfetal growth.d.People living on the island of Palau have the highest rates of schizophrenia in the world.H0:The betelnut that is chewed on the island causes schizophrenia.e.Orangutans living in zoos tend to be overweight when compared to their wildcounterparts.H0:Orangutans living in zoos do not exercise as much as those living in the wild and arefed a steady diet, causing them to be overweight compared to their wild counterparts whomust forage for their food.Exercise 6a.Review an article from a biological anthropology journal. Below list the titlesandfunctions of the varioussections.*Answers may vary depending on journal article chosen*•For example, Abstract–summary of paper•Introduction–what the study is going to do, background literature•Materials and Methods–how theresearcher(s)performed the experiment and whatsamples they used to do so•Results–the results of the experiment, usually with lots of tables and graphs•Discussion–what the results mean and how theyfitwithinthe existing knowledge of thesubject•Conclusion–A summing up of the data and how this information will be used and addedto in the futureb.In which section do you find the hypothesis being tested, or the study questions?Introduction

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c.Where would you look to find the details necessary to repeat this experiment?Materials and MethodsPost-Lab Questions1.How does modern science differ from faith?How do these compare interms of teleologicalor cause-and-effect explanations?Modern science is based on scientific methodology and the ability to test the possibleanswers to natural phenomena. Faith is based on belief without needing to “prove”or testsuch beliefs. Modern science relies upon cause and effect, while faith relies upon teleology2.Describe the three assumptions all sciences are based on.Natural Causality–All phenomena have a basis in natureor natural explanationUniformity in Space and Time–All phenomena will occur the same way regardless of thewhenever or wherever it occursCommon Perception–all individuals perceive events through their senses in the same way3.Describe the steps of the scientific method.Observation–A researcher observes a natural phenomena that is repeatableHypothesis–The researcher comes up with a reason why the natural phenomena occursExperiment-The researcher tests his/her hypothesis to see if it is accurateConclusion–The researcher sums up the experiment and determines if the hypothesis wasaccurate or not.4.What does an experiment test?The hypothesis5.Describe the difference between the:•independent variable–the variable thatadjusted/manipulated to test the outcome

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•dependent variable-thevariablethat is being tested, it may be affected bychanging the independent variable•control variable–Variables that are being held constant by the researcher6.Using Exercise 2, 3, or4, can you identify any variables that were not controlled forin thestudythat could have beencontrolled?*Answers will vary and may include:*Exercise 2–Amount of sleep,genetic predispositions,other medications and supplements…Exercise 3–Amount of exercise, amount of light…Exercise 4–size of the pot the plants are in, type of soil7.Reviewing your team’s answers to Exercise5, can you think of an alternate test of yourhypothesis?Was anything left out of your original experiment?Student answers will vary.8.Choose a research article from a physical anthropology journal online or in the library.Canyou identify what hypothesis the author(s) is/are testing?Is the experiment designed in such away that it might be repeated by another investigator?Student answers will vary.

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Chapter 2: Cell Biology and DNAReadingQuestions1.Which of the following individuals was responsible for coining the term “cell”?(a) Hooke,(b) Darwin, (c) Wilkins, (d) Watson.2.Prokaryotic cells are distinguishable from eukaryotic cells because prokaryotes donotcontain: (a) organelles, (b) a plasma membrane, (c) DNA,(d) a nucleus.3.Chromosome strands are called (a) centromeres, (b) alleles,(c) chromatids, (d) homologues.4.Alternate forms of a gene are called(a) alleles, (b) sister chromatids (c) homologues (d)replicated DNA.5.Sister chromatids separate during nuclear division in (a) mitosis, (b) meiosis I, (c) meiosis II,(d) both a and c.6.Who won the Nobel Prize in 1962 for identifying the structure of DNA? (a) Hooke, (b)Meischer, (c) Watson and Franklin,(d) Watson, Crick, and Franklin.7.Which of the following is a possible base pairing in DNA? (a) adenine-cytosine,(b)adenine-thymine, (c) cytosine-thymine, (d) thymine-guanine.8.Transcription in DNA (a) results in the formation of an identical DNA strand,(b) results inthe formation of mRNA, (c) happens in the nucleus, (d) requires the assistance of tRNAanticodons.9.True orFalse: DNA replication occurs in the ribosome.10.Trueor False: Crossing over is an important source of variability.In-Class ExercisesExercise 1Why do gametes have only 23 chromosomes, one of each pair?Because when they combineduring fertilization the zygote will have the correct number of chromosomes, one of each pairfrom mom and the other of each pair from dad.Chimpanzees have 48 chromosomes in their somatic cells. How many chromosomes do youthink are found in their sex cells?24

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Exercise 2*Exercise requires karyotype and answers will vary*Exercise 3*Exercise requires microscope and answers will vary depending on phases present*Exercise 4Compare and contrast mitosis and meiosis in the human with the following matching questions.a1. happens in the bodycellsa. mitosisb2. produces 4 daughter cellsc3. begins with 46 chromosomesb. meiosisa4. produces 2 daughter cellsa5. one nuclear divisionc. both mitosis and meiosisc6. one chromosome replicationb7. happens in the testes and ovariesb8. daughter cells have 23 chromosomes eachb9. two nuclear divisionsa10. daughter cells are diploidExercise 5Draw a homologous pair of chromosomes.Use one color (e.g.pink) for one member ofthe pair and use a second color (e.g. blue) for the second member of the pair.Next, draw the two chromosomes crossing over, so that the two colors are touching.Third, draw the two chromosomes after the crossing over is completed and they have shuffledtheir gene pairs, exchanging genes (colors) between them.Have at least one exchange.Compareyour drawing to others in the class and see the amount of variation that might be possible.*Answers will vary but may look like the following:*

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Exercise6Practice DNA base pairing:Consider the followingDNA strand:A T C C T A G G T C A GIdentify the complementary bases:T A G G A T C C A G T CNow, practice DNA replication.Consider the following double stranded DNA molecule.Notice that the DNA bases are paired accordingly.Separate the strands and replicate them,identifying which strands are original and which are the new complementary strands.Write thecomplementary bases for the top strand above the strand and the complementary bases for thebottom strand below the strand.New strand:A T G C C G T T G A C T C G ATop strand:T A C G G C A A C T G A G C T_______________________________________Bottom strand:A T G C C G T T G A C T C G ANew strand:T A C G G C A A C T G A G C TExercise 7

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The following chart lists all possible mRNA codons and the 20 amino acidsthey codefor. Note that there is some redundancy in the code.Also note that some codons code for start orstop, which tells the cell where to start or stop making the protein.Using this information, fill inthe blanks below the chart forthe amino acideach codon calls for.UCASerineGUAValineUGGTryptophanAGAArginineCUCLeucineGCCAlanineCAUHistidineAUGStart (Methionine)Exercise 8The following is atemplate strand of DNA:A C G G T T C A T G C Aa.What is the complimentary mRNA strand?U G C C A A G U A C G Ub.What are the complimentary tRNA anticodons?ACG; GUU; CAU; GCAc.Using the chart from the previous exercise, what is the sequence of amino acids for thispeptide chain?Be sure to use the mRNA codons when reading the chart!UGC:Cysteine;CAA:Glutamine;GUA:Valine;CGU:ArgininePost-Lab Questions1.Describe the difference between the autosomes and the sex chromosomes.Answers should include some or all of the following:Autosomes are always homologouspairs and contain information pertaining to body structure and function; they comprise pairs1-22 in humans. The sex chromosomes(pair 23)are homologous in females but not inmales, since the X and Y chromosome are different lengths and the Y chromosome carriesinformationprimarily pertaining to the biological sex of the individual.

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2.How many chromosomes were there in your karyotype set? Was this the normal number forhumans?*Depends upon the student’s karyotype*. 46is the normal number.Some kits offermutations.3.Referring to your lecture textbook, or theInternet, discuss the clinical symptoms associatedwith any anomaly you identified in your karyotype.*Student activity, answers will vary.*Some karyotype kits offer Turner’s syndrome,Klinefelter’s syndrome, Down Syndrome, and XYY Males.4.How do you determine the sex of an individual when examining their karyotype?By looking at the sex chromosomes–if there are 2 identical chromosomes, the individual isXX and female. If one of the chromosomes is small and the other is large, the individual isXY and male.5.How are the different types of chromosomes identified for a karyotype?Based on size, length of arms and position of the centromere.6.If the chromosome number for an organism is 22 before mitosis, what is the chromosomenumber of each daughter cell after mitosis has taken place?227.WhydoesDNA replicate prior to mitosis?So that each daughter cell has the complete complement of chromosomes.8.What do you think might happen ifa cell underwent mitosis but not cytokinesis?The cell without the cytoplasm and associated organelles would not survive(this is commonin females, one gamete gets all cellular contents, the ovum, while the other three getlittle/none and are called polar bodies, which resorb).

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9.If a cell in an organism had 16 chromosomes before meiosis, how many chromosomes wouldexistin each nucleus after meiosis? What is the diploid number?What is the haploidnumber?8after meiosis, 16 is diploid, 8 is haploid.10.From a genetic standpoint, what is the significance of fertilization?It is when the egg and sperm meet allowing the 23 chromosomes from the mother to unitewith the 23 chromosomes from the father (in humans) creating a zygote.11.Describe the differences between haploid and diploid cells. Where arehaploid and diploidcellsfound?Diploid cells have the full complement of chromosomes with all the homologous pairs.Haploid cells have only half of the complement of chromosomes, with only one of eachchromosomefrom each homologous pair.Body cells are diploid, sex cells are haploid.12.Discuss the differences you observed when comparing your crossing over diagram to othersin the class.How many different combinations did you see?*Student activity–answers will vary because each person’s diagram will be at least slightlydifferent.*13.What does it mean when we say DNA replication is semiconservative?Oneparental strand remains intact, while a new complementary strand is formed.14.Describe the differences in DNA and RNA structure.DNA is double stranded while RNA is single stranded. DNA has a deoxyribose sugar, RNAhas ribose sugar; DNA has thymine, RNA does not, but has Uracil.15.To transcribe means “make a copy of”.Is an exact copy of DNA made duringthe process oftranscription?Why or why not?No, because RNA does not have thymine, so it replaces itwith Uracil. Also, because of the law of complementary bases, the RNA strand is actually a

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“mirror image” of the DNA strand that is being copied.Also the sugar molecules differbetween DNA and RNA.16.Where does transcription happen?What about translation?Transcription happens in the nucleus, translation occurs in the ribosome.17.What amino acid would be produced iftranscription took place from the DNA sequenceCAT?(mRNA would be GUA),amino acid isValine.•If a genetic mistake took place during replication and the new DNA strand has thesequenceCAG,what is the new mRNA, andwhichamino acid would this result in?(mRNA is GUC), amino acidisValine.•What if the genetic mistake resulted in a DNA strand with the sequence GAT?(mRNA is CUA),amino acid isLeucine.•Explain these results.Because there is some redundancy in the codons versus aminoacids some genetic mistakes will not result in a change of amino acids, while othermistakes will.

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Chapter 3:Principlesof InheritanceReadingQuestions1.Mendel published his works in (a) 1965, (b) 1956, (c) 1776,(d) 1865.2.In a dihybrid cross (a) only one trait is considered, (b) the parents are always heterozygous,(c) two traits are considered, (d) all offspring must be heterozygous.3.When all self fertilized offspring display the same traits as their parents, we know that theparents are (a) hybrids, (b) heterozygous, (c) codominant,(d) true breeders.4.The physical characteristics of an organism are referred to as the (a) dominant allele, (b)gametes,(c) phenotype, (d) genotype.5.An individual who is carrying two of the same alleles for a gene is known as(a)homozygous,(b) heterozygous, (c) dominant, (d) a hybrid.6.The allele that is masked or hidden in the genotype is the (a) heterozygote,(b) recessiveallele, (c) dominant allele, (d) true breeder.7.When both alleles are fully expressed in the phenotype, this is called (a) incompletedominance, (b) codominance, (c) sex-linked, (d) recessive.8.Trueor False: Sex-linked traits are often located on the X chromosome.9.True orFalse: Mendel’s Law of Independent Assortment states that during meiosis, thechromosomes pair separates, so that each newly formed gamete receives one chromosomefrom each pair.10.Trueor False: Numerous traits in humans are inherited in a simple Mendelian fashion.

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In-Class ExercisesNo number/text question.•Can you explain this discovery using our modern understanding of cell biology and DNAfrom Chapter 2?The particles that Mendel referred to were genes. Each parent contributed one gene for eachtrait, the same as would happen in meiotic division to form the gametes and the subsequentformation of the zygote. Mendel is describing meiosis and zygote formation.Exercise 1a.Mendel’s pea plants carry two alleles for the flower color gene, P for purple flowers, andp for white flowers. What three possible combinations might exist in any one plant?PP, Pp, ppb.Mendel’s pea plants also carry two genes for plant height, T for tallplants and t for shortplants.Consider the genotypes in thefollowing tableand indicate the possible genotypesof the gametes (Remember: Each gamete will only carry one allele).Diploid genotypeGamete genotypeGamete genotypeTTTTTtTtttTtc.Working in the opposite direction, during fertilization when the sperm and egg fuse, thehaploid cells come together and form a diploid zygote.In the table below, provide thediploid genotypes that would occur by fusion of the following gamete genotypes.Gamete genotype(Sperm)Gamete genotype(Ovum)Diploid genotype(Zygote)TtttTtTtTTTT

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Exercise 2We know thatpurple flowers in pea plantsare dominant to white flowers.Using the example forflowercolor above in Exercise 1, identify the flowercolors for plantsthat have the followinggenotypes.Label the homozygous and heterozygous conditions.*PPHomozygous dominant-Purple*PpHeterozygous-Purple*ppHomozygous recessive-WhiteWhy?If a dominant allele is present, regardless of what the other allele is, the phenotype will bedominant.The dominant allele masks the presence of the recessive allele. In order for arecessive allele to be expressed, there must be two copies of the recessive allele.Exercise 3In Mendel’s pea plants, yellow seeds are dominant to green seeds.Using Y for yellow and y forgreen, list the three possible genotypes, followed by their phenotype (yellow or green), and labelthe homozygous and heterozygous conditions.•YY–Homozygous dominant–Yellow•Yy–Heterozygous–Yellow•Yy–Homozygous recessive-GreenExercise 4For this exercise, let’s expand our practice to include three traits at once.a.What gametes are produced from a plant that is Pp?P or pb.What gametes are produced from a plant that is PpTt?PT, Pt, pT, or ptc.What gametes are produced from a plant that is PpTtYy?PTY, Pty, PtY, Pty, pTY, pTy,ptY, or ptyExercise 5

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a.Considering that purple is dominant to white, and using P for purple and p for white,draw a Punnett square crossing a heterozygous purple plant and a white flowered plant.PppPppppPpppb.What are the genotypes produced by this mating?Pp and ppc.What are the phenotypes produced by this mating?Pp–purple and pp-whited.What percentage of the offspring is homozygous?50%Exercise 6a.We know that the allele for round seeds is dominant to the allele for wrinkled seeds. Using Rfor round and r for wrinkled,draw a Punnett squarecrossing two plants that are bothheterozygous.RrRRRRrrRrrrb.What percentage of the offspring is homozygous dominant?25%c.What percentage of the offspring is homozygous recessive?25%d.What percentage of the offspring is heterozygous?50%e.What percentage of the offspring is wrinkled?25%f.What percentage of the offspring is round?75%Why?Because both the heterozygous andthe homozygous dominant will have the round phenotype.Exercise 7a.Calculate the Punnett square crossing two pink flowered petunias.

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RWRRRRWWRWWWWhat percentage of the offspring is pink?50%White?25%Red?25%b.Calculate the Punnett square crossing a roan shorthorn bull with a white cow.RWWRWWWWRWWWWhat percentage of the offspring has a roan coat?50%A red coat?0%A white coat?50%Exercise 8a.Using X for normal vision and XCfor colorblind, calculate the Punnett square that wouldresult from a carrier femalemating with a colorblind male.Hint: the female genotype isXCX and the male genotype is XCY.XcXXcXcXcXcXYXcYX Yb.What are the chances that they will produce a colorblind son?(1 out of 2) 50%c.What are the chances that they will produce a colorblind daughter?(1 out of 2) 50%d.Is there any way these parents might producea daughter with normal vision?YesIf so,how?If the mother givesher “normal” X to her daughter and the father donates his onlyaffected X, then the daughter will be XcX(normal vision, but a carrier).e.Is there any chance these parents might produce a son with normal vision?YesIf so,how?If the mother gives her “normal” X to her son,and dad donated his Y,then the sonwould be XY.

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Exercise 9a.Draw a Punnett square crossing a heterozygous taster and a nontaster.PppPppppPpppb.What percentage of the offspring is homozygous dominant?0%c.What percentage of the offspring is homozygous recessive?50%d.What percentage of the offspring are tasters?50%Exercise 10a.Draw a Punnett Square crossing a homozygous dominant person and a heterozygousperson.EEEEEEEeEeEeb.What are the chances of getting a child with the flaky gray earwax?0%c.What percentage of the offspring is heterozygous?50%d.What percentage of the offspring is homozygous?50%Exercise11For each trait in the chart below list your phenotype.Using the information collected byeveryone inyour class, fill in the rest of the chart.*Student activity–no answers provided.This chart will be different for each student and eachclass!*Exercise 12a.List all of the possible genotypes for an individual with the ability to roll the tongue and acleft chin (hint: there are four possible combinations).TTDD; TtDD; TTDd; TtDd

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b.List all possible combinations for an individual that cannot roll his or hertongue and hasa cleft chin.ttDD; ttDdc.List all possible genotypes for an individual that can roll his or hertongue and does nothave a cleft chin.TTdd; Ttddd.What is the genotype for an individual who cannot rollhis or hertongue and does nothave a cleft chin?ttdde.Suppose an individual is heterozygous for both traits(tongue rolling and cleft chin). Whatis their genotype?TtDdf.What are the genotypes of the gametes this person could produce?TD; Td; tD; tdExercise 13Using the answer you got in Exercise 12, questionf, you can set up a Punnett square for adihybrid problem,crossing two individuals that are heterozygous for tongue rolling and cleftchin.This square will have 16 boxes.Insert the parentalgenotypes on the top and left side of thebox below, and calculate the possible offspring genotypes.When you are finished answer thequestions below.TDTdtDtdTDTdtDtda.What are the chances of having a child that is:*TTDD=1/16=0.0625*TtDd=4/16 =0.25*TtDD=2/16 =0.125*ttDd=2/16=0.125b.What are the chances of having a child that has a cleft chin and cannot roll their tongue?TTDDTTDdTtDDTtDdTTDdTTddTtDdTtddTtDDTtDdttDDttDdTtDdTtddttDdttdd

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3/16=0.1875c.What are the chances of having a child that does not have a cleft chin and can roll theirtongue?3/16 =0.1875d.What are the chances of having a child that cannot roll their tongue and does not have acleft chin?1/16=0.0625Exercise 14Cystic fibrosisis the mostcommon metabolic error in European-derived (“white”)populations,where about 1of every 1,600 beinga carrier.Individuals born with cystic fibrosis are lacking anenzyme that allows them to break down thick mucus in the lungs and makes them susceptible toserious and often fatal lung infections. With aggressive treatment, mostindividuals may reachadulthood;otherwise death from pneumonia is likely in childhood.There is no cure;however,genetic engineering offers promise for these affected individuals.Cystic fibrosis is inherited as arecessive, and individuals with one copy of the gene are carriers of the disease.The gene hasbeen located on chromosome 7.a.What percentage of the gametes of a heterozygote individual will contain the recessiveallele?50%b.Draw a Punnett square crossing two individuals who are heterozygous for this trait.(Using F for fibrosis, and f for no fibrosis)FfFFFFffFfffc.What percentage of the offspring will be affected with cystic fibrosis?25%d.What percentage of the offspring will be normal?75%e.What percentage of the offspring will also carry the trait?50%f.Is it possible for an affected child to be born to one healthy parent and one carrier?No,both parents must carry the gene.

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Post-Lab Questions1.Distinguish between incomplete dominance and codominance.Codominant alleles are both fully expressed in the phenotype(e.g. red + white = redandwhite); Incomplete dominant alleles have a “blended” appearance in the phenotype(e.g. red+ white = pink).2.What does it mean when we say that a trait is sex-linked?It is on one of the sex chromosomes, usually the X chromosome.3.How might the Law of Independent Assortment be violated if two traits were on the samechromosome?Genes found on the same chromosome, especially if they are near each other (e.g. on thesame arm) will usually be inherited together.4.Mendel’s pea plants carry two alleles of the genefor seed shape, R for round seeds and r forwrinkledseeds.Consider the following genotypes in the table below, and indicate thepossible genotypes of the gametes.Diploid genotypeGamete genotypeGamete genotypeRRRRRrRrrrrr5.In the chart below, provide the diploid genotypes that would occur by fusion of the followinggenotypes.Gamete genotypeGamete genotypeDiploid genotyperrrrRrRrRRRR

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6.Forthe followingproblem, state whether the genotype is homozygous and heterozygous andwhether theplant would have round or wrinkled seeds.*RRHomozygous dominant-round*RrHeterozygous-round*rr Homozygousrecessive-wrinkled7.Considering that the allele for tall plants is dominant to the allele for short plants, and using Tfor tall and t for short, draw a Punnett square crossing a heterozygous tall plant and a shortplant.TttTttttTttt*Whatpercentage of the offspring ishomozygous dominant?0%*Whatpercentage of the offspring isheterozygous?50%*Whatpercentage of the offspring ishomozygous recessive?50%*What percentage of the offspring istall?50%Why?Any plant with a T will be tall*Whatpercentage of the offspring isshort?50%Why?Plants with two t (tt) will be short8.Considering that the allele in humans for tongue rolling is dominant to the allele for non-rolling and using R for rolling and r for non-rolling, draw a Punnett Square crossing aheterozygous roller and a homozygous roller.RRRRRRRrRrRr*What percentage of the offspring is homozygous dominant?50%*What percentage of the offspring is heterozygous?50%*What percentage of the offspring is homozygous recessive?0%*What percentage of the offspringcan roll their tongue?100%Why?Any plant with a Rwill be a roller

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*What percentage of the offspringcannot roll their tongue?0%Why?No individualsarerr9.Considering that the allele for a dimpled chin is dominant to the allele for a non-dimpledchin, and using D for dimpled and d for non-dimpled, draw a Punnett square crossing ahomozygous person with a dimpled chin and a person without a dimpled chin.DDdDdDddDdDd*What percentage of the offspring is homozygous dominant?0%*What percentage of the offspring is heterozygous?100%*What percentage of the offspring is homozygous recessive?0%*What percentage of the offspring has a dimpled chin?100%Why?Anypersonwith a Dwill be dimpled*What percentage of the offspring does not have a dimpled chin?0%Why?Noindividualsaredd10.Working with the dihybrid cross you did in Exercise 13answer the following questions:*What are the chances of having child that is ttdd?1/16 0.0625*What are the chances of having child that is ttDD?1/16 0.0625*What are the chances of having child that is Ttdd?2/16 0.125*What are the chances of having a child thathas a cleft chin and can roll their tongue?9/16 0.562511.Using the chart below, copy your personalphenotypeinformation collected in class in thefirst columnand ask or examine your parents for their phenotype for each trait.Try anddetermine your genotype. (If you are unable to determine your parents’phenotype,you mayask a friend orrelative for their information).*Student activity, no answer provided.This question will be answered differently for eachstudent.*

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Chapter 4: Human Variation: Blood Groups and Pedigree AnalysisReadingQuestions1.An example of codominance in the ABO blood group system is: (a) AO, (b) BB,(c) AB, (d)OO.2.Type A blood has:(a) A antigens, (b) B antigens, (c) neither A nor B antigens, (d) Oantigens.3.Type O blood has which of the following possible genotypes? (a) AO, (b) BO,(c) OO, (d)none of the above.4.Individuals with a disorder that is inherited as a recessive will express the trait if they carry:(a) two copies of the allele, (b) one copy of the allele, (c) the allele on their sexchromosome, (d) the allele on the X chromosome.5.In dominant inheritance, when one parent expresses the trait and the other does not, theaffected individual must be: (a) homozygous dominant, (b) homozygous recessive,(c)heterozygous,(d) a female.6.Females expressing sex-linked recessive traits like red-green colorblindness: (a) areheterozygous, (b) must carry the trait on both of their X chromosomes, (c) only need tohave one x affected to express the trait, (d) express the trait when their only x is affected.7.When a male’s single X chromosome is affected by a sex-linked trait, that male isconsidered: (a) homozygous recessive, (b) homozygous dominant, (c) heterozygous,(d)hemizygous.8.Trueor False: Under the ABO blood group system, a universal donor may give blood toanyone.9.True orFalse: It is possible for dominant traits to skip a generation.10.Trueor False: Individuals affected with a recessive condition may have phenotypicallynormal parents.

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In-Class ExercisesExercise 1a.Can you list which of the genotypes above are homozygous?AA, BB, OOb.Can youlist which of the genotypes above are heterozygous?AO, BO, ABExercise 2a.Can a person with type A blood receive a blood transfusion from a person with type O?Yesb.Can a person with type B blood receive a blood transfusion from a person with type AB?Noc.Can a person with type O blood donate blood to a person who is type AB?Yesd.Can a person with type B blood donate blood to a person who is type O?NoExercise 3You will be given a sample of artificial blood. Use the kit provided to determine the blood typeof your sample and answer the following questions.*Student activity–answers willvary for each student.*Exercise 4A woman who is blood type B gives birth to a child who is blood type O. Using this information,answer the questions and fill in the genotypes and phenotypes in the following table (part e).a.What is the mother’s genotype?BOb.The mother claims a certain man who is blood type B is the father of this child. Is thispossible? If so how, if not, why not?Yes, If he is BOc.This woman has a second child who is blood type A. Is it possible that the same man isthe father of this child?No, the mother must donate her O allele and the father musthave an A allele (AA or AO), which the proposed father does not have if he is has type Bblood.d.There is a third child in the house whose blood type is AB. The woman claims this is hersister’s child, also fathered by the same man. Her sister’s blood type is O. What does this

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suggest to you?While the mancouldbe the father of this child, the woman’s sister couldnot be the mother. An individual who is type O cannot have a child who is type AB.e.These two sisters have a brother who is blood type O. Based on the information you haveabout these three siblings, determine the possible blood types of their parents.GenotypePhenotypeMotherBOBSisterOOOBrotherOOOChild #1OOOChild #2AOAChild #3ABABProposed FatherBOBGrandmother*BO*BGrandfather*?O*A, B, or O*The Grandmother and Grandfather could be reversed.Exercise 5Practice drawing a simple pedigree. Sharon and John are married.They have three children, twoboys(Justin and Ian)and one girl(Dana).Justin is married with one daughter.Danais marriedwith one daughter.Sharon is affected by a dominant trait, which is also seen in Justin and hisdaughter.No other family members are affected.Writethe genotypes for each member of thefamily above their symbol.Sharon: DdJohn: ddIan: ddJustin: DdJustin’s Wife: ddJustin’s daughter: DdDana: ddDana’s Husband: ddDana’s daughter: ddExercise 6The followingis a sample pedigree for familial hypercholesterolemia. Note that allaffected individuals have at least one parent who isalso affected by the disease.Working ingroups, or guided by your instructor, determine the genotypes for each individual in the pedigree.You may write the answers below the symbols.

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*Note: Pedigree uses D for the dominant allele and d for the recessive allele.Trait is dominant.May also use H for dominant and h for recessive.*Exercise 7The followingis a sample pedigree fora family affected by albinism.Note that most affectedindividuals have parentswho are phenotypically normal and thatthe parentsmust be carriers ofthe trait.Working in groups, or guided by your instructor, determine the genotypes for eachindividual in the pedigree.*Note:Pedigree uses Rfor the dominant alleleand rfor the recessive allele. Trait isrecessive.May also use A for dominant and afor recessive.*

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Exercise 8Using thesample pedigree exhibiting hemophilia in the British royal family(Stanford etal. 2006), working in groups, or guided by your instructor, determine the genotypes for eachnamedindividual in the pedigree. Remember to label the females as XX and the males as XY.Affected X chromosomes may be notated as Xh.Albert: XYVictoria: XXhEmpress Victoria: XXKaiser Wilhelm II: XYEdward VII: XYGeorge V: XYAlice of Hesse: XXhPrincess Irene: XXhFredrick: XhY(Alexandra) Alix: XXhHelena Princess Christian: XXLeopold Duke of Albany: XhYAlice of Athlone: XXhBeatrice: XXhVictoria Eugenie: XXhLeopold: XYMaurice: XhYPost-Lab Questions1.Using the information from Exercise 3, and the blood typing you did in class, can youdetermine at least a partial genotype for the parents of the individual whose blood you tested?Yes–(it will depend upon the answer to Exercise 3, but at least a partial can be determinedfor all blood types)

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