unit overview genetics - temecula valley unified …web1.tvusd.k12.ca.us/gohs/myoung/bio....

What You’ll LearnChapter 10

Mendel and Meiosis

Chapter 11DNA and Genes

Chapter 12Patterns of Heredity and Human Genetics

Chapter 13Genetic Technology

Unit 4 ReviewBioDigest & Standardized Test Practice

Why It’s ImportantPhysical traits, such as the stripes of these tigers, areencoded in small segments of a chromosome called genes,which are passed from one generation to the next. Bystudying the inheritance pattern of a trait through severalgenerations, the probability that future offspring willexpress that trait can be predicted.

1863Lincoln writes the EmancipationProclamation.

1865Mendel discovers the rulesof inheritance.

Understanding the PhotoWhite tigers differ from orange tigers by having ice-blueeyes, a pink nose, and creamy white fur with brown orblack stripes. They are not albinos. The only time a whitetiger is born is when its parents each carry the white-coloring gene. White tigers are very rare, and today, theyare only seen in zoos.

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Genetics

(t)Science Photo Library/Photo Researchers, (crossover)Tom Brakefield/CORBIS

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The following standards are covered in Unit 4:Investigation and Experimentation: 1a, 1d, 1i, 1j, 1k, 1mBiology/Life Sciences: 2a, 2b, 2c, 2d, 2e, 3, 3a, 3b, 3c, 3d, 4a, 4b, 4c, 5a, 5b,5c, 5d, 5e, 7b, 7c

California Standards

Genetics

Unit OverviewUnit 4 presents genetics and itsrole in determining traits oforganisms. Chapter 10 introduces geneticsthrough a historical presentationof Gregor Mendel’s work. Meiosisis then introduced. Chapter 11 presents the struc-ture of DNA and replication isexplained. Transcription andtranslation are explained. Variousmutations are described. Chapter 12 examines non-Mendelian heredity and the prin-ciples of human genetics. Chapter 13 discusses selectivebreeding, DNA technology, andthe Human Genome Project.

Introducing the UnitHave students look at the pictureof the tiger cubs and describe thetraits they see. The white tiger isnot an albino, or a separatespecies. It is a tiger with recessivecolor genes. Unit 4 will discusshow genes determine an organ-ism’s physical traits, as well as howthose traits are inherited.

A WebQuest isan inquiry-based online projectin which all information used bystudents is obtained from theWeb. Students evaluate the infor-mation to complete the activity.Access the Unit 4 WebQuest at

Discussion As students read Unit 4, have them refer to the abovetime line. Ask students to infer howMendel’s discovery of the rules ofinheritance (1865) led to the conclu-sion that DNA is the genetic material

(1952) and how both events led tothe Human Genome Project (1990s).Have student groups research and dis-cuss future uses of the humangenome map.

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ca.bdol.glencoe.com/webquest

http://ca.bdol.glencoe.com/webquest

http://ca.bdol.glencoe.com/webquest

1950The Korean War beginswhen North Koreainvades South Korea.

1961The genetic codeis cracked.

2000Most of thehuman DNAsequence iscompleted.

1990The Human GenomeProject begins to mapand sequence the entirehuman genome.

1952Alfred Hershey andMartha Chase showconclusively that DNAis the genetic material.

1910Scientists determine that genesreside on chromosomes.

1944Scientists suggest geneticmaterial is DNA, not protein.The results are not accepted.

1964The Beatles maketheir first appearanceon American TV.

1953Watson, Crick,Wilkins, andFranklin determinethe structure of DNA.

(tl)Omikron/Science Source/Photo Researchers, (tr)CNRI/Phototake, NYC

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Advance MaterialsPlanningThe following materials may needto be ordered a few weeks inadvance of the planned activity.

Chapter 10■ MiniLab (p. 254) microscope

slide and coverslip, microscope■ Project (p. 257) Brassica rapa

seeds■ MiniLab (p. 268) clay■ BioLab (p. 274) small flower-

pots or seedling flats

Chapter 11■ Additional Lab (p. 282) 3-ply

yarn■ Quick Demo (p. 285) zipper■ Quick Demo (p. 298) coiled

telephone cord

Chapter 12■ Additional Lab (p. 316) het-

erozygous tobacco seeds■ Project (p. 321) mustard seeds■ Additional Lab (p. 324) pre-

pared slides of male andfemale cheek cells

■ BioLab (p. 330) Brassica rapaseeds—normal and variegated

Chapter 13■ MiniLab (p. 350) Random

page from a novel ■ Additional Lab (p. 350) pre-

pared slide of cheek cells,human hair (blond and dark)

■ Quick Demo (p. 351) humanchromosome map

Using the LibraryIntrapersonal Research one diseasein which gene therapy is an experi-mental therapeutic approach suchas cancer, hemophilia, Alzheimer’sdisease, and sickle cell anemia.

InterviewLinguistic Interview someone witha family pedigree that demon-strates an inherited disease such asHuntington’s disease, hemophilia,and Tay-Sachs disease.

Final ReportHave student groups present theirfindings about genetics in reportsthat could be presented to stu-dents at your local middle school.

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1950The Korean War beginswhen North Koreainvades South Korea.

1961The genetic codeis cracked.

2000Most of thehuman DNAsequence iscompleted.

1990The Human GenomeProject begins to mapand sequence the entirehuman genome.

1952Alfred Hershey andMartha Chase showconclusively that DNAis the genetic material.

1910Scientists determine that genesreside on chromosomes.

1944Scientists suggest geneticmaterial is DNA, not protein.The results are not accepted.

1964The Beatles maketheir first appearanceon American TV.

1953Watson, Crick,Wilkins, andFranklin determinethe structure of DNA.

Section ObjectivesNational Science State/Local Advanced Lab and

Standards Standards Demo Planning

End of Chapter AssessmentStudent Edition

Study Guide, p. 277Content Assessment, pp. 278–279

2 sessions, 2 blocks

1. Relate Mendel’s two laws to theresults he obtained in his experi-ments with garden peas.

2. Predict the possible offspring of agenetic cross by using a Punnettsquare.

3 sessions, 2 blocks

3. Analyze how meiosis maintains aconstant number of chromosomeswithin a species.

4. Infer how meiosis leads to varia-tion in a species.

5. Relate Mendel’s laws of heredityto the events of meiosis.

UCP.1–3, UCP.5;A.1, A.2; G.1–3

UCP.1–3; A.1, A.2;C.1, C.2; E.1, E.2;F.6; G.1–3

Student Labs:MiniLab 10.1, p. 254: flower, water,glass slide, coverslip, pencil with eraser,microscopeProblem-Solving Lab 10.1, p. 262

Teacher Demonstration:Quick Demo, p. 259: 300 blackbeans, 100 white beans, small cup or bag

Student Labs: Problem-Solving Lab 10.2, p. 264Additional Lab, p. 266: long length ofyarn, paper clips, long length of string,toothpicks, tape or glue, scissorsMiniLab 10.2, p. 268: clayInternet BioLab, p. 274: See materials below.

Teacher Demonstration:Quick Demo, p. 266

Student Lab:Internet BioLab, p. 274: potting soil,small flowerpots or seedling flats, twogroups of tobacco seeds, hand lens, lightsource, thermometer, plant-watering bottle

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Level 1 activitiesshould be appropriatefor students withlearning difficulties.

Level 2 activitiesshould be within theability range of all students.

Level 3 activitiesare designed forabove-average students.

ELL activitiesshould be within theability range ofEnglish LanguageLearners.

Cooperative Learningactivities are designedfor small group work.

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Biology/LifeSciences2c, 2g, 3, 3a, 3bInvestigation &Experimentation1j, 1k

Biology/LifeSciences2a, 2b, 2c, 2d, 2e,3b, 3d*Investigation &Experimentation1i

Reproducible Masters Technology

and Transparencies

Unit 4 FAST FILE ResourcesMiniLab Worksheet, p. 3BioLab Worksheet, pp. 5–6Real World BioApplications, pp. 7–8Reinforcement and Study Guide in English,

pp. 9–10Reinforcement and Study Guide in Spanish,

pp. 13–14Critical Thinking/Problem Solving, p. 18Transparency Worksheets, pp. 19, 21–22, 25–26

Reading Essentials for Biology, Section 10.1

Section Focus Transparency 24Basic Concepts Transparency 14Reteaching Skills Transparency 16

Unit 4 FAST FILE ResourcesMiniLab Worksheet, p. 4Reinforcement and Study Guide in English,

pp. 11–12Reinforcement and Study Guide in Spanish,

pp. 15–16Concept Mapping, p. 17Transparency Worksheets, pp. 20, 23–24, 27–28

Reading Essentials for Biology,Section 10.2

Laboratory Manual, pp. 53–54, 55–58Section Focus Transparency 25Basic Concepts Transparency 15Reteaching Skills Transparency 17

Unit 4 FAST FILE ResourcesChapter Assessment, pp. 29–34Student Recording Sheet, p. 35

Reviewing Biology, pp. 19–20

Interactive Chalkboard CD-ROM: Section 10.1 PresentationTeacherWorks™ CD-ROMGuided Reading Audio Summaries MP3Virtual Labs CD-ROMVirtual Lab: Punnett Squares

Interactive Chalkboard CD-ROM: Section 10.2 PresentationTeacherWorks™ CD-ROMGuided Reading Audio Summaries MP3

Interactive Chalkboard: Chapter 10 AssessmentMindJogger Videoquizzes DVD/VHSExamView® Pro Test Bank CD-ROM TeacherWorks™ CD-ROM

Transparency CD-ROM MP3 Videocassette DVD

Legend

/self_check_quiz/vocabulary_puzzlemaker/chapter_test/standardized_test

Succeeding on National Standards CD-ROM

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ca.bdol.glencoe.com

Indicates materials created specifically for California.

http://ca.bdol.glencoe.com

Short on Time?

The BioDigest at theend of this unit can be used as a(n):• preview to introduce important

unit concepts.• overview if time does not per-

mit teaching the entire chapter.• review of key unit concepts.

Understandingthe PhotoA zebra’s stripes are thought to bean adaptation that helps themavoid predators. Zebras recognizeeach other by their stripes. Theirstripes are like fingerprints—theyare passed on genetically, andevery zebra’s stripe pattern is different.

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What You’ll Learn■ You will identify the basic

concepts of genetics.■ You will examine the process

of meiosis.

Why It’s ImportantGenetics explains why you haveinherited certain traits fromyour parents. If you understandhow meiosis occurs, you can seehow these traits were passed onto you.

Mendel and MeiosisMendel and Meiosis

Dr. G.G. Dimijian/Photo Researchers

Visit to• study the entire chapter

online• access Web Links for more

information and activities ongenetics and meiosis

• review content with theInteractive Tutor and self-check quizzes

Zebras usually travel in largegroups, and each zebra’s stripesblend in with the stripes of thezebras around it. This confusespredators. Rather than seeingindividual zebras, predators seea large, striped mass. Zebrastripe patterns are like humanfingerprints—they are geneti-cally determined, and everyzebra’s stripe pattern is unique.

Understandingthe Photo

ca.bdol.glencoe.com

Demo Show students wrinkled andsmooth pea seeds. Have them com-pare and contrast the physical traits ofthe pea seeds. Encourage them tospeculate about less obvious traits

such as biochemical or physiologicalcharacteristics. The wrinkled nature ofthe seeds is due to a lack of well-formed starch grains and lower waterretention. L2

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10.1SECTION PREVIEWObjectivesRelate Mendel’s two lawsto the results he obtainedin his experiments withgarden peas.Predict the possible off-spring of a genetic cross byusing a Punnett square.

Review Vocabularyexperiment: a procedure

that tests a hypothesisby the process of col-lecting data under con-trolled conditions (p. 13)

New Vocabularyhereditytraitgeneticsgametefertilizationzygotepollinationhybridalleledominantrecessivelaw of segregationphenotypegenotypehomozygousheterozygouslaw of independentassortment

heredity from theLatin word hered-,meaning “heir”;Heredity describesthe way the gene-tic qualities youreceive from yourancestors arepassed on.

Heredity Make the following Foldable to help you organize information about Mendel’s laws of heredity.

Fold one piece of paper lengthwise into thirds.

Fold the paper widthwise into fifths.

STEP 1 STEP 2

MendelDescribe inYour Words

Give anExample

Rule ofunit factors

Rule of dominance

Law ofsegregation

Law of independentassortment

Make a Table As you read Chapter 10, complete the table describingMendel’s rules and laws of heredity.

Unfold, lay the paper length-wise, and draw lines along the folds.

Label your table as shown.

STEP 3 STEP 4

10.1 MENDEL’S LAWS OF HEREDITY 253

Mendel’s Laws of Heredity

Why Mendel SucceededPeople have noticed for thousands of years that family resemblances are

inherited from generation to generation. However, it was not until themid-nineteenth century that Gregor Mendel, an Austrian monk, carriedout important studies of heredity—the passing on of characteristics fromparents to offspring. Characteristics that are inherited are called traits.Mendel was the first person to succeed in predicting how traits are trans-ferred from one generation to the next. A complete explanation requiresthe careful study of genetics—the branch of biology that studies heredity.

Mendel chose his subject carefullyMendel chose to use the garden pea in his experiments for several rea-

sons. Garden pea plants reproduce sexually, which means that they producemale and female sex cells, called gametes. The male gamete forms in thepollen grain, which is produced in the male reproductive organ. Thefemale gamete forms in the female reproductive organ. In a process calledfertilization, the male gamete unites with the female gamete. The resultingfertilized cell, called a zygote (ZI goht), then develops into a seed.

Standard 3b Students know the genetic basis for Mendel’s laws ofsegregation and independent assortment.


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BIOLOGY: The Dynamics of Life SECTION FOCUS TRANSPARENCIES

Use with Chapter 10,Section 10.1

What possible combinations can result from combiningone coin from each group?

What is the ratio of the possible combinations?

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Unit 4 FAST FILE ResourcesMiniLab Worksheet, p. 3BioLab Worksheet, pp. 5–6Real World BioApplications, pp. 7–8Reinforcement and Study Guide in

English, pp. 9–10Reinforcement and Study Guide in

Spanish, pp. 13–14Critical Thinking/Problem Solving, p. 18

Transparency Worksheets, pp. 19,21–22, 25–26

Section Focus Transparency 24Basic Concepts Transparency 14Reteaching Skills Transparency 16 Reading Essentials for Biology,

Section 10.1

1 FocusBellringerSection Focus Transparency 24

This CD-ROM is an editableMicrosoft® PowerPoint®

presentation that includes:• Section presentations• Section checks• Image bank• Hot links to Biology Online• All transparencies

FOLDABLES™For an additional Foldablesactivity idea, see the Chapter 10Foldables page in Unit 4 FAST

FILE Resources.

Figure 10.1 In his experiments,Mendel often had totransfer pollen fromone plant to anotherplant with differenttraits. This is called making a cross.Describe How didMendel make a cross?

The transfer of pollen grains from amale reproductive organ to a femalereproductive organ in a plant is calledpollination. In peas, both organs arelocated in the same flower and aretightly enclosed by petals. This pre-vents pollen from other flowers fromentering the pea flower. As a result, peasnormally reproduce by self-pollination;that is, the male and female gametescome from the same plant. In many ofMendel’s experiments, this is exactlywhat he wanted. When he wanted tobreed, or cross, one plant with another,Mendel opened the petals of a flowerand removed the male organs, as shownin Figure 10.1A. He then dusted thefemale organ with pollen from the planthe wished to cross it with, as shown inFigure 10.1B. This process is calledcross-pollination. By using this tech-nique, Mendel could be sure of the par-ents in his cross.

Mendel was a careful researcherMendel carefully controlled his

experiments and the peas he used. Hestudied only one trait at a time to con-trol variables, and he analyzed his datamathematically. The tall pea plants heworked with were from populations ofplants that had been tall for manygenerations and had always producedtall offspring. Such plants are said tobe true breeding for tallness. Like-wise, the short plants he worked withwere true breeding for shortness.

Removemale parts

Transferpollen

Female part Male parts

Cross-pollination

Pollengrains

AA BB

254 MENDEL AND MEIOSIS

Observe and InferLooking at Pollen Pollen grainsare formed within the maleanthers of flowers. What istheir role? Pollen contains themale gametes, or sperm cells,needed for fertilization. Thismeans that pollen grains carrythe hereditary units from maleparent plants to female parentplants. The pollen grains thatMendel transferred from theanther of one pea plant to thefemale pistil of another plant carried the hereditary traits that he so carefully observed in the next generation.

Procedure! Examine a flower. Using the diagram as a guide, locate the

stamens of your flower. There are usually several stamensin each flower.

@ Remove one stamen and locate the enlarged end—theanther.

# Add a drop of water to a microscope glass slide. Place theanther in the water. Add a coverslip. Using the eraser endof a pencil, tap the coverslip several times to squash theanther.

$ Observe under low power. Look for numerous small roundstructures. These are pollen grains.

Analysis1. Estimate Provide an estimate of the number of pollen

grains present in an anther.2. Describe What does a single pollen grain look like?3. Explain What is the role of pollen grains in plant

reproduction?

Pistil

Anther

Stamens

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2 Teach

Purpose Students will observe pollengrains in plants.

Process Skillsobserve and infer

Safety PrecautionsCaution students when handlingslides and coverslips. Have students wash their hands afterhandling flowers and flower parts.

Teaching Strategies■ Preserved flower material is

available from biological sup-ply houses.

■ Forceps may be used to re-move stamens from the flower.

■ If coverslips or slides breakduring the squashing process,have students place them in acontainer for broken glass andstart over with new material.

Expected ResultsStudents will observe that eachanther contains thousands ofsmall pollen grains.

Analysis1. Numbers should be in the

several thousands.2. Student answers will vary

depending on species used.Pollen grains are microscopiccells, often round in shape.

3. Pollen grains contain malereproductive cells needed forfertilization. They providechromosomes and genes ofthe male parent.

AssessmentPerformance Have students dia-gram several pollen cells underhigh power and have them includepollen size, in micrometers. If stu-dents have not done any previousmeasuring, refer them to MiniLab7.1. Use the Performance TaskAssessment List for Using Mathin Science in PASC, p. 101. L2

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Caption Question AnswerFigure 10.1 Mendel removedpollen from the anther of oneflower and placed it on the stig-ma of another.

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Mendel’s MonohybridCrosses

What did Mendel do with the talland short pea plants he selected? Hecrossed them to produce new plants.Mendel referred to the offspring ofthis cross as hybrids. A hybrid is theoffspring of parents that have differ-ent forms of a trait, such as tall andshort height. Mendel’s first experi-ments are called monohybrid crossesbecause mono means “one” and thetwo parent plants differed from eachother by a single trait—height.

The first generationMendel selected a six-foot-tall pea

plant that came from a population ofpea plants, all of which were over sixfeet tall. He cross-pollinated this tallpea plant with pollen from a short peaplant that was less than two feet talland which came from a population ofpea plants that were all short. Whenhe planted the seeds from this cross,he found that all of the offspring grewto be as tall as the taller parent. In thisfirst generation, it was as if theshorter parent had never existed.

The second generationNext, Mendel allowed the tall

plants in this first generation to self-pollinate. After the seeds formed, heplanted them and counted more than1000 plants in this second generation.Mendel found that three-fourths ofthe plants were as tall as the tall plantsin the parent and first generations.He also found that one-fourth of theoffspring were as short as the shortplants in the parent generation. Inother words, in the second genera-tion, tall and short plants occurred ina ratio of about three tall plants to oneshort plant, as shown in Figure 10.2.The short trait had reappeared as iffrom nowhere.

The original parents, the true-breeding plants, are known as the P1generation. The P stands for “parent.”The offspring of the parent plants are known as the F1 generation. The F stands for “filial”—son ordaughter. When you cross two F1plants with each other, their offspringare the F2 generation—the second fil-ial generation. You might find it eas-ier to understand these terms if you


Short pea plant

All tall pea plants

3 tall : 1 short

Tall pea plant

F2

F1

P1

Figure 10.2 When Mendel crossed true-breeding tall pea plantswith true-breeding short pea plants, all the off-spring were tall. When he allowed first-generationtall plants to self-pollinate, three-fourths of the off-spring were tall and one-fourth were short.

EnrichmentLinguistic Some researchers havedoubted Mendel’s honesty,despite the accuracy of his con-clusions. One claim is thatMendel deliberately excludeddata from traits that do not inde-pendently assort because he hadonly one chance in 6000 of ran-domly selecting one gene on eachof seven chromosomes. Anotherclaim is that Mendel fabricatedsome of his data in order to makethem fit expected ratios.

Countering these claims areother analyses of Mendel’s datathat concluded he actually had aone in three chance of choosingonly traits that independentlyassort, and a suggestion thatMendel intended to presentsomething like a demonstrationrather than an exact experiment.Have interested students researchthis topic and draw their ownconclusions in a report for theirportfolios. L3

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Visually Impaired: Kinesthetic Obtaina large flower model. Allow thosestudents who are visually impaired tohandle the model. Direct their atten-tion to the stamen and pistil. Then,have students examine an actualflower after having studied themodel.

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Mendel For those students needingan additional challenge, have themdraw a concept map of Mendel’smonohybrid cross through the second generation. They shouldinclude which traits were dominantand which were recessive. Post wellorganized concept maps around theclassroom. L3

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Pages 254–255: Biology/Life Sciences 2c

look at your own family. Your parentsare the P1 generation. You are the F1generation, and any children youmight have in the future would be theF2 generation.

Mendel did similar monohybridcrosses with a total of seven pairs oftraits, studying one pair of traits at atime. These pairs of traits are shownin Figure 10.3. In every case, hefound that one trait of a pair seemedto disappear in the F1 generation,only to reappear unchanged in one-fourth of the F2 plants.

The rule of unit factorsMendel concluded that each

organism has two factors that controleach of its traits. We now know that these factors are genes and thatthey are located on chromosomes.Genes exist in alternative forms. We call these different gene forms alleles (uh LEELZ). For example, each

of Mendel’s pea plants had two allelesof the gene that determined itsheight. A plant could have two allelesfor tallness, two alleles for shortness,or one allele for tallness and one forshortness. An organism’s two allelesare located on different copies of achromosome—one inherited fromthe female parent and one from themale parent.

The rule of dominance Remember what happened when

Mendel crossed a tall P1 plant with a short P1 plant? The F1 offspringwere all tall. In other words, only one trait was observed. In such crosses,Mendel called the observed trait dominant and the trait that disap-peared recessive. Mendel concludedthat the allele for tall plants is domi-nant to the allele for short plants.Thus, plants that had one allele fortallness and one for shortness were tall.


allele from theGreek wordallelon, meaning“of each other”;Genes exist inalternative formscalled alleles.

round yellow purpleaxial(side) green inflated tall

wrinkled green whiteterminal

(tips) yellow constricted short

Seedshape

Dominanttrait

Recessivetrait

Seedcolor

Flowercolor

Flowerposition

Podcolor

Podshape

Plantheight

Figure 10.3 Mendel chose seventraits of peas for hisexperiments. Each traithad two clearly differ-ent forms; no interme-diate forms wereobserved. CompareWhat genetic varia-tions are observed in plants?

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Caption Question AnswerFigure 10.3 Mendel studiedvariations in seed shape andcolor, flower color and position,pod color and shape, and plantheight. Many other variationsalso exist.

Rules of Heredity For students needingan additional challenge, have themwork in groups and imagine that theyare Gregor Mendel and that they havejust formulated the principles of domi-nance and segregation. Have them

work together to write an illustratedarticle describing their findings for alocal newspaper. They should remem-ber that the year is 1863. SeeBioChallenges and Enrichment for addi-tional projects and activities. L3

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Almost all students think thatthe dominant trait must bemost common or desirable.However, the dominant traitis merely one that masks analternate allele that may bepresent in the heterozygouscondition.

Uncover theMisconceptionAsk students to give examplesof dominant traits. When astudent names a recessivetrait that is really dominant,point out the error.

Demonstrate theConceptExplain that Huntington’sdisease, polydactyly, andhypercholesterolemia (hav-ing dangerously high levelsof blood cholesterol) arethree dominant traits inher-ited in humans, although allthree traits are very rare.

Assess NewKnowledgeHave students state the defi-nition of dominance. Askthem to state how it is possi-ble for a dominant trait torarely appear. The number ofrecessive alleles in the genepool for a particular traitmay be far greater than thenumber of dominant allelesin the pool.

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Expressed another way, the allele forshort plants is recessive to the allele fortall plants. Pea plants with two allelesfor tallness were tall, and those withtwo alleles for shortness were short.You can see in Figure 10.4 how therule of dominance explained theresulting F1 generation.

When recording the results ofcrosses, it is customary to use the sameletter for different alleles of the samegene. An uppercase letter is used forthe dominant allele and a lowercaseletter for the recessive allele. Thedominant allele is always written first.Thus, the allele for tallness is writtenas T and the allele for shortness as t, asit is in Figure 10.4.

Describe Mendel’stwo rules of heredity.

The law of segregationNow recall the results of Mendel’s

cross between F1 tall plants, whenthe trait of shortness reappeared. Toexplain this result, Mendel formu-lated the first of his two laws ofheredity. He concluded that each tallplant in the F1 generation carriedone dominant allele for tallness andone unexpressed recessive allele forshortness. Each plant received theallele for tallness from its tall parentand the allele for shortness from itsshort parent in the P1 generation.Because each F1 plant has two differ-ent alleles, it can produce two typesof gametes—“tall” gametes and“short” gametes. This conclusion,illustrated in Figure 10.5 on the nextpage, is called the law of segregation.The law of segregation states thatevery individual has two alleles ofeach gene and when gametes areproduced, each gamete receives oneof these alleles. During fertilization,these gametes randomly pair to pro-duce four combinations of alleles.

Dwight R. Kuhn

Tall plant

a

All tall plantsT t

Short plantT T t

tT

t

F1

Figure 10.4 The rule of dominance explains theresults of Mendel’s cross between P1 talland short plants (A). Tall pea plants areabout six feet tall, whereas short plantsare less than two feet tall (B).

AA

BB

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Visual LearningFigure 10.4 Have students writeout the three important writtenconventions that are describedwith this diagram. Use the sameletter for different alleles of thesame gene; use uppercase let-ters for dominant alleles andlowercase letters for recessivealleles; and always write thedominant allele first.

The rule ofunit factors says that thereare two different factorsthat control each trait. Therule of dominance explainsthat one of the two factors(dominant allele) governs atrait and the alternate formof the trait (recessive) is onlypresent if the dominant fac-tor is not present.

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Experimental Crosses Rapid-cyclingBrassica rapa plants complete their lifecycle in 30 to 40 days. They are idealfor use in the classroom because withinthis short time, the plants flower andform seeds for the next generation.Genetic studies can be carried out using

different traits, such as petal-less flow-ers or hairy stems. Students will find iteasy to carry out the pollinatingbetween plants. These plants are anideal experimental organism for genet-ic studies. L2

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Pages 256–257: Biology/Life Sciences 2c, 2g, 3a, 3bInv. & Exp. 1k

Phenotypes andGenotypes

Mendel showed that tall plants arenot all the same. Some tall plants,when crossed with each other, yieldedonly tall offspring. These wereMendel’s original P1 true-breedingtall plants. Other tall plants, whencrossed with each other, yielded bothtall and short offspring. These werethe F1 tall plants in Figure 10.5 thatcame from a cross between a tall plantand a short plant.

Two organisms, therefore, can lookalike but have different underlyingallele combinations. The way anorganism looks and behaves is calledits phenotype (FEE noh tipe). Thephenotype of a tall plant is tall,whether it is TT or Tt. The allele

combination an organism contains isknown as its genotype ( JEE noh tipe).The genotype of a tall plant that hastwo alleles for tallness is TT. Thegenotype of a tall plant that has oneallele for tallness and one allele forshortness is Tt. You can see that anorganism’s genotype can’t always beknown by its phenotype.

An organism is homozygous (hohmoh ZI gus) for a trait if its two allelesfor the trait are the same. The true-breeding tall plant that had two allelesfor tallness (TT) would be homozy-gous for the trait of height. Becausetallness is dominant, a TT individual is homozygous dominant for that trait. A short plant would always have twoalleles for shortness (tt). It would,therefore, always be homozygousrecessive for the trait of height.


phenotype fromthe Greek wordsphainein, meaning“to show,” andtypos, meaning“model”; The visi-ble characteristicsof an organismmake up its phenotype.genotype fromthe Greek wordsgen or geno,meaning “race,”and typos, mean-ing “model”; Theallele combinationof an organismmakes up its genotype.

Tall plant

3 1

TallT T

TallT t

TallT t

Shortt t

Tall plant

F1

F2

Law of segregation Tt ×× Tt cross

T t T t

Figure 10.5 Mendel’s law of segregationexplains the results of hiscross between F1 tall plants.He concluded that the twoalleles for each trait mustseparate when gametes areformed. A parent, therefore,passes on at random only oneallele for each trait to eachoffspring.

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Using Science TermsIt is sometimes easier for studentsto remember the meaning of theterm genotype by reversing theword so it becomes “type ofgene.” When stated this way, stu-dents can associate the term geno-type with its definition. Phenomeans “to show.” Thus, the phe-notype shows the type of trait orhow it appears.

Concept Development■ Ask students to supply the cor-

rect term—genotype or phe-notype—to the followingexamples: (a) LL, (b) blondhair, (c) dimpled chin, (d) blueeyes, (e) Dd, (f ) ss, (g) whiteand green leaves. a, e, and fare genotypes; b, c, d, and gare phenotypes.

■ Have students provide the fol-lowing information for thisexample: G � green pea pod, g � yellow pea pod. (a) Give thephenotypes of plants with thesegenotypes: Gg, GG, and gg. (b)Use the terms homozygous or heterozygous to describe each ofthe three examples above. (a) green, green, yellow; (b)heterozygous, homozygous,homozygous

Have the students research tofind and define other wordswith “homo” as part of the root,such as homogenous and homol-ogous.

Selective Breeding: KinestheticSelective breeding of plants is crucialto agriculture. Students can try this forthemselves by using fast-growingplants and selecting for a desired trait.During this activity, students will haveto plant and care for their plants aswell as perform the cross-pollinations.

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An organism is heterozygous (hehtuh roh ZI gus) for a trait if its twoalleles for the trait differ from eachother. Therefore, the tall plant thathad one allele for tallness and oneallele for shortness (Tt) is heterozy-gous for the trait of height.

Now look at Figure 10.5 again.Can you identify the phenotype andgenotype of each plant? Is each planthomozygous or heterozygous? Youcan practice determining genotypesand phenotypes in the BioLab at theend of this chapter.

Mendel’s DihybridCrosses

Mendel performed another set ofcrosses in which he used peas that dif-fered from each other in two traitsrather than only one. Such a crossinvolving two different traits is calleda dihybrid cross because di means“two.” In a dihybrid cross, will thetwo traits stay together in the nextgeneration or will they be inheritedindependently of each other?

The first generationMendel took true-breeding pea

plants that had round yellow seeds(RRYY) and crossed them with true-breeding pea plants that had wrinkledgreen seeds (rryy). He already knewthat when he crossed plants that pro-duced round seeds with plants thatproduced wrinkled seeds, all theplants in the F1 generation producedseeds that were round. In otherwords, just as tall plants were domi-nant to short plants, the round-seeded trait was dominant to thewrinkled-seeded trait. Similarly, whenhe crossed plants that produced yellow seeds with plants that producedgreen seeds, all the plants in the F1 generation produced yellow seeds—yellow was dominant. Therefore,

Mendel was not surprised when hefound that the F1 plants of his dihy-brid cross all had the two dominanttraits of round and yellow seeds, asFigure 10.6 shows.

The second generationMendel then let the F1 plants polli-

nate themselves. As you might expect,he found some plants that producedround yellow seeds and others thatproduced wrinkled green seeds. Butthat’s not all. He also found someplants with round green seeds andothers with wrinkled yellow seeds.When Mendel sorted and countedthe plants of the F2 generation, hefound they appeared in a definiteratio of phenotypes—9 round yellow:3 round green: 3 wrinkled yellow: 1 wrinkled green. To explain theresults of this dihybrid cross, Mendelformulated his second law.


heterozygousfrom the Greekwords heteros,meaning “other,”and zygotos,meaning “joinedtogether”; A traitis heterozygouswhen an individualhas two differentalleles for thattrait.

P1

F1

F2

Round yellow Wrinkled green

All roundyellow

9Roundyellow

3Roundgreen

3Wrinkledyellow

1Wrinkled

green

Dihybrid cross round yellow ×× wrinkled green

Figure 10.6 When Mendel crossed true-breeding plants that produced round yellowseeds with true-breeding plants that produced wrinkled green seeds, theseeds of all the offspring were round and yellow. When the F1 plantswere allowed to self-pollinate, they produced four different kinds ofplants in the F2 generation.

The BioLab atthe end of thechapter can beused at thispoint in the lesson.

Inherited Traits Mix 300black beans with 100 whitebeans. Give each student asmall cup or bag of beans.Explain that the color of thebeans represents an inherit-ed trait. Ask students tospeculate as to which trait isdominant and what thegenotypes for color wouldbe for the beans. The blackcolor is dominant and theblack beans would be eitherBB or Bb. The white beanswould be bb. L2

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English Language Learners To help stu-dents predict genes within each gametefor a dihybrid cross and to illustrate thelaw of independent assortment, providethem with the mnemonic: 1 and 3 and 1 and 4, 2 and 3 and 2 and 4. Have stu-dents write the numbers 1–4 across thetop of parental alleles in a dihybrid cross.

1 2 3 4R r Y y

To obtain the correct gamete combina-tions according to the law of independ-ent assortment, students should matchalleles 1 and 3 (RY), 1 and 4 (Ry), 2 and 3(rY), 2 and 4 (ry).

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ResourcesFor additional high-interestactivities, see the Forensics andBiotechnology Lab Manual.

The law of independent assortment

Mendel’s second law states thatgenes for different traits—for exam-ple, seed shape and seed color—areinherited independently of eachother. This conclusion is known asthe law of independent assortment.When a pea plant with the genotypeRrYy produces gametes, the alleles Rand r will separate from each other(the law of segregation) as well asfrom the alleles Y and y (the law of

independent assortment), and viceversa. These alleles can then recom-bine in four different ways. If the alle-les for seed shape and color wereinherited together, only two kinds ofpea seeds would have been produced:round yellow and wrinkled green.

Punnett SquaresIn 1905, Reginald Punnett, an

English biologist, devised a shorthandway of finding the expected propor-tions of possible genotypes in the off-spring of a cross. This method iscalled a Punnett square. It takesaccount of the fact that fertilizationoccurs at random, as Mendel’s law ofsegregation states. If you know thegenotypes of the parents, you can usea Punnett square to predict the possi-ble genotypes of their offspring.

Monohybrid crossesConsider the cross between two F1

tall pea plants, each of which has thegenotype Tt. Half the gametes of eachparent would contain the T allele, andthe other half would contain the t allele. A Punnett square for thiscross is two boxes tall and two boxeswide because each parent can producetwo kinds of gametes for this trait.The two kinds of gametes from oneparent are listed on top of the square,and the two kinds of gametes fromthe other parent are listed on the leftside, as Figure 10.7A shows. It doesn’tmatter which set of gametes is on topand which is on the side, that is,which parent contributes the T alleleand which contributes the t allele.Refer to the Punnett square in Figure 10.7B to determine the possi-ble genotypes of the offspring. Eachbox is filled in with the gametes aboveand to the left side of that box. Youcan see that each box then containstwo alleles—one possible genotype.


Heterozygoustall parent

Heterozygoustall parent

T

T

T t

tt

T t

Figure 10.7 This Punnett square predicts the results of amonohybrid cross between two heterozygouspea plants.

T

TT Tt

Tt

T

t

t tt

You can see that thereare three differentpossible genotypes—TT, Tt, and tt—and thatTt can result from twodifferent combinations.Interpret ScientificIllustrations Howmany possiblephenotypes resultfrom this cross?

B

The gametes thateach parent formsare listed on thetop and left side of the Punnettsquare.

A

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Concept DevelopmentLogical-Mathematical Have stu-dents construct Punnett squaresand solve these problems, givingthe genotypic and phenotypicratios expected. (a) Homozygoustall plant bred to a homozygousshort plant. All offspring will betall and heterozygous. (b)Homozygous tall plant bred to aheterozygous tall plant. All off-spring will be tall, half being TTand half being Tt. (c) Hetero-zygous tall plant bred to a homozy-gous short plant. Half will be talland half will be short; all tall off-spring will be Tt and all short off-spring will be tt.

Biology JournalPunnett Square Have studentsconstruct a Punnett square toillustrate the probable bean par-ents from the Quick Demo on theprevious page. Ask students towrite a prediction of what theoutcome would be if two whitebeans were crossed.

Caption Question AnswerFigure 10.7 Two

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Candy Genes Provide students with twolarge and two small round pieces ofcandy plus two large and two small but-tons. Advise them that the buttons andcandy represent two different genes andthe large-sized objects represent domi-nant alleles. Ask them to prepare twosets of objects (alleles) for two traits intwo parents, making both parents

heterozygous for both traits. Then havestudents arrange their parental gene setsinto independently assorted gametes oftwo alleles each. Advise them that theycannot have two candies or two buttonsin their final groups. Have them explainhow this illustrates the law of independ-ent assortment.

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CD-ROM Students will predict thetraits of offspring in Punnett Square.

Pages 260–261: Biology/Life Sciences 2c, 2g, 3a, 3b

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After the genotypes have beendetermined, you can determine thephenotypes. Looking again at thePunnett square in Figure 10.7B, youcan see that three-fourths of the off-spring are expected to be tall becausethey have at least one dominant allele.One-fourth are expected to be shortbecause they lack a dominant allele.Of the tall offspring, one-third will behomozygous dominant (TT) and two-thirds will be heterozygous (Tt).Note that whereas the genotype ratiois 1TT: 2Tt: 1tt, the phenotype ratio is3 tall: 1 short. You can practice doingcalculations such as Mendel did in theConnection to Math at the end of thischapter.

Dihybrid crossesWhat happens in a Punnett square

when two traits are considered?Think again about Mendel’s crossbetween pea plants producing roundyellow seeds and plants producingwrinkled green seeds. All the F1plants produced seeds that wereround and yellow and were heterozy-gous for each trait (RrYy). What kindof gametes will these F1 plants form?

Mendel explained that the traits forseed shape and seed color would beinherited independently of eachother. This means that each F1 plantwill produce gametes containing thefollowing combinations of genes withequal frequency: round yellow (RY),round green (Ry), wrinkled yellow(rY), and wrinkled green (ry). APunnett square for a dihybrid crosswill then need to be four boxes oneach side for a total of 16 boxes, asFigure 10.8 shows.

ProbabilityPunnett squares are good for show-

ing all the possible combinations ofgametes and the likelihood that each

will occur. In reality, however, youdon’t get the exact ratio of resultsshown in the square. That’s because,in some ways, genetics is like flippinga coin—it follows the rules of chance.

When you toss a coin, it landseither heads up or tails up. The prob-ability or chance that an event willoccur can be determined by dividingthe number of desired outcomes bythe total number of possible out-comes. Therefore, the probability ofgetting heads when you toss a coinwould be one in two chances, writtenas 1:2 or 1⁄2. A Punnett square can be used to determine the probabilityof getting a pea plant that producesround seeds when two plants that are heterozygous (Rr) are crossed.


Gametes from RrYy parentRY

RRYY RRYy RrYY RrYy

RRYy RRyy RrYy Rryy

RrYY RrYy rrYY rrYy

RrYy Rryy rrYy rryy

Ry rY ry

RY

Ry

rY

ry

round yellow

round green

wrinkled yellow

wrinkled green

F1 cross: RrYy × RrYy

Punnett Square of Dihybrid Cross

Gam

etes

from

RrY

y pa

rent

Figure 10.8 A Punnett square for a dihybrid cross betweenheterozygous pea plants producing round yel-low seeds shows clearly that the offspring ful-fill Mendel’s observed ratio of 9 round yellow:3 round green: 3 wrinkled yellow: 1 wrinkledgreen.

Check for UnderstandingHave students describe therelationship between oramong the following terms.

a. pollination—fertilization b. allele—dominant—

recessive c. genotype—phenotype d. homozygous—

heterozygous e. monohybrid—dihybrid

ReteachHave students provide anexample of each relationshipprovided in Check forUnderstanding.

ResourcesFor more practice, useReading Essentials for Biology,Section 10.1.

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Visual LearningFigure 10.8 Ask students howmany different genotypes and phe-notypes resulted from this cross. 9 genotypes and 4 phenotypes

3 Assess

ExtensionHave students list the genotypesand phenotypes resulting from (a)an RrYy plant cross-pollinated byan RRyy plant; (b) an rrYy plantcross-pollinated by a RrYy plant.(a) genotypes: RrYy, RRyy, RRYy,Rryy; phenotypes: 1/2 round yel-low, 1/2 round green; (b) geno-types: RrYY; RrYy, Rryy, rrYY, rrYy,rryy; phenotypes: 3 round yellow;3 wrinkled yellow; 1 roundgreen; 1 wrinkled green

Knowledge Have students ex-plain what each of the followingrepresents in a Punnett square: (a)the letters written at the top andside of the square; (b) the letterswritten within each box; (c) theboxes. (a) gametes; (b) genotypeof an offspring; (c) possible off-spring L1

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Learning Disabled/English LanguageLearners: Visual-Spatial Provide stu-dents with blank Punnett square out-lines. Project a similar copy onto a screenwith an overhead projector. Lead stu-dents through the steps showing a crossbetween two Tt parents. Reinforce howthe letters placed to the side and across

the top represent all the possiblegametes for each parent and how thisillustrates the law of segregation.Reinforce the significance of the foursquares and what the letter combina-tions within them represent. Provide stu-dents with a variety of problems (TT �tt, Tt � tt, Tt � TT, tt � tt).

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Understanding Main Ideas1. What structural features of pea plant flowers

made them suitable for Mendel’s genetic studies?

2. What are the genotypes of a homozygous and aheterozygous tall pea plant?

3. One parent is homozygous tall and the other isheterozygous. Make a Punnett square to showhow many offspring will be heterozygous.

4. How many different gametes can an RRYy parentform? What are they?

Thinking Critically5. In garden peas, the allele for yellow peas is domi-

nant to the allele for green peas. Suppose you have

a plant that produces yellow peas, but you don’tknow whether it is homozygous dominant or het-erozygous. What experiment could you do to findout? Draw Punnett squares to help you.

6. Observe and Infer The offspring of a crossbetween a plant with purple flowers and a plantwith white flowers are 23 plants with purple flow-ers and 26 plants with white flowers. Use the letter P for purple and p for white. What are thegenotypes of the parent plants? Explain your rea-soning. For more help, refer to Observe and Inferin the Skill Handbook.

SKILL REVIEWSKILL REVIEW


R

R

RR

Rr

Rr

rr

r

r

Figure 10.9The probability that the offspring from amating of two heterozygotes will show adominant phenotype is 3 out of 4, or 3/4.

The Punnett square in Figure 10.9shows three plants with round seedsout of four total plants, so the proba-bility is 3⁄4. Yet, if you calculate theprobability of round-seeded plantsfrom Mendel’s actual data in theProblem-Solving Lab on this page, you will see that slightly less thanthree-fourths of the plants wereround-seeded. It is important toremember that the results predictedby probability are more likely to beseen when there is a large number ofoffspring.

Analyze InformationData Analysis In addition to crossing tall and short pea plants, Mendel crossed plants that formed round seeds with plants that formed wrinkled seeds. He found a 3:1 ratio of round-seeded plants to wrinkled-seeded plants in the F2 generation.

Solve the ProblemMendel’s F2 results are shown to the right.1. Calculate the actual

ratio of round-seeded plants to wrinkled-seeded plants by dividing the number of round-seeded plants by the number of wrinkled-seededplants. Your answer tells you how many more times round-seeded plants resulted than wrinkled-seeded plants.

2. To express your answer as a ratio, write the number fromstep 1 followed by a colon and the numeral 1.

Thinking Critically1. Compare How does Mendel’s observed ratio compare

with the expected 3:1 ratio? 2. Analyze Why did the actual and expected ratios differ?

RR

rr

Mendel’s Results

Kind of NumberPlant of Plants

Round-seeded 5474

Wrinkled-seeded 1850

ca.bdol.glencoe.com/self_check_quiz

PurposeStudents will convert and expresstwo sets of related numbers as aratio.

BackgroundThe round seeds contain an abun-dance of well-formed starch grainsand have higher water retentionthan the wrinkled seeds. A lack ofwell-formed starch grains andlower water retention account forthe wrinkled appearance.

Process Skillsuse and interpret data, calculate

Teaching Strategies■ Have students work in groups.

Place students with poor mathskills with students working atgrade level.

■ Provide students with an exam-ple if they are having difficultyin determining a ratio.

Thinking Critically1. The observed ratio is slightly

lower than expected. 2. The observed and expected

ratios may differ slightly dueto chance.

AssessmentSkill Provide students with otherexamples of traits and data. Havethem calculate ratios. Use thePerformance Task AssessmentList for Using Math in Science inPASC, p. 101. L2

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1. They are self-pollinating, and maleflower parts can be easily removed toallow for cross-pollination.

2. homozygous tall � TT, heterozygous �Tt

3. One-half of all offspring will be het-erozygous. (Note: It will not matter if

the homozygous parent is homozy-gous dominant or recessive.)

4. two; RY and Ry5. Cross the unknown yellow plant with

a recessive green parent. If the off-spring are all yellow, the unknowngenotype is homozygous yellow. If

half the offspring are yellow and halfgreen, the unknown genotype is het-erozygous.

6. Genotypes of parents are Pp for the purple-flowered plant and pp forthe white-flowered plant. Purple isdominant.

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10.2SECTION PREVIEWObjectivesAnalyze how meiosismaintains a constant num-ber of chromosomeswithin a species.Infer how meiosis leads tovariation in a species.Relate Mendel’s laws ofheredity to the events ofmeiosis.

Review Vocabularymitosis: the orderly

process of nuclear divi-sion in which two newdaughter cells eachreceive a complete setof chromosomes (p. 204)

New Vocabularydiploidhaploidhomologous chromosomemeiosisspermeggsexual reproductioncrossing overgenetic recombinationnondisjunction

10.2 MEIOSIS 263

Meiosis

Biophoto Associates/Photo Researchers

Solving the PuzzleUsing an Analogy Mendel’s studyof inheritance was based oncareful observations of peaplants, but pieces of thehereditary puzzle were stillmissing. Modern tech-nologies such as high-power microscopes allowus a glimpse of things thatMendel could only imag-ine. Chromosomes, such asthose shown here, were themissing pieces of the puzzlebecause they carry the traitsthat Mendel described. The keyto solving the puzzle was discoveringthe process by which these traits aretransmitted to the next generation.Organize Information As you read this section, make a list of the ways in which meiosis explains Mendel’s results.

Metaphase chromosomes

Color-enhanced SEM Magnification: 650�

Genes, Chromosomes, and NumbersOrganisms have tens of thousands of genes that determine individual

traits. Genes do not exist free in the nucleus of a cell; they are lined up onchromosomes. Typically, a chromosome can contain a thousand or moregenes along its length.

Diploid and haploid cellsIf you examined the nucleus in a cell of one of Mendel’s pea plants,

you would find it had 14 chromosomes—seven pairs. In the body cells ofanimals and most plants, chromosomes occur in pairs. One chromosomein each pair came from the male parent, and the other came from thefemale parent. A cell with two of each kind of chromosome is called adiploid cell and is said to contain a diploid, or 2n, number of chromo-somes. This pairing supports Mendel’s conclusion that organisms havetwo factors—alleles—for each trait. One allele is located on each of thepaired chromosomes.

Organisms produce gametes that contain one of each kind of chromo-some. A cell containing one of each kind of chromosome is called a haploidcell and is said to contain a haploid, or n, number of chromosomes.

Standard 2a Students know meiosis is an early step in sexual reproductionin which the pairs of chromosomes separate and segregate randomly during cell division to product gametes containing one chromosome of each type.


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BIOLOGY: The Dynamics of Life SECTION FOCUS TRANSPARENCIES

Use with Chapter 10,Section 10.2

What is the number of chromosomes in each body cell ofthese fruit flies?

How many chromosomes must each body cell of normaloffspring have?

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SECTION FOCUS

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Unit 4 FAST FILE ResourcesMiniLab Worksheet, p. 4Reinforcement and Study Guide in

English, pp. 11–12Reinforcement and Study Guide in

Spanish, pp. 15–16Concept Mapping, p. 17Transparency Worksheets, pp. 20,

23–24, 27–28

Section Focus Transparency 25Basic Concepts Transparency 15Reteaching Skills Transparency 17Reading Essentials for Biology,

Section 10.2Laboratory Manual, pp. 53–54, 55–58

1 FocusBellringerSection Focus Transparency 25

Using an AnalogyOrganize Information Meiosisexplains the results of simpledominance experiments, as wellas the principles of segregationand independent assortment.

Pages 262–263: Biology/Life Sciences 2c, 2e, 2g, 3,3a Inv. & Exp. 1j

This fact supports Mendel’s conclusionthat parent organisms give one factor,or allele, for each trait to each of theiroffspring.

Each species of organism contains acharacteristic number of chromo-somes. Table 10.1 shows the diploidand haploid numbers of chromosomes

of some species. Note the large rangeof chromosome numbers. Note alsothat the chromosome number of aspecies is not related to the complexityof the organism.

Homologous chromosomesThe two chromosomes of each pair

in a diploid cell are called homologous(hoh MAH luh gus) chromosomes.Each of a pair of homologous chro-mosomes has genes for the sametraits, such as plant height. On homo-logous chromosomes, these genes arearranged in the same order, butbecause there are different possiblealleles for the same gene, the twochromosomes in a homologous pairare not always identical to each other.Identify the homologous chromo-somes in the Problem-Solving Lab.

Let’s look at the seven pairs ofhomologous chromosomes in the cellsof Mendel’s peas. These chromosomepairs are numbered 1 through 7. Eachpair contains certain genes located atspecific places on the chromosome.Chromosome 4 contains the genes forthree of the traits that Mendel studied.Many other genes can be found onthis chromosome as well.

Every pea plant has two copies ofchromosome 4. It received one fromeach of its parents and will give one atrandom to each of its offspring.Remember, however, that the twocopies of chromosome 4 in a pea plantmay not necessarily have identicalalleles. Each chromosome can haveone of the different alleles possible foreach gene. The homologous chromo-somes diagrammed in Figure 10.10show both alleles for each of threetraits. Thus, the plant represented bythese chromosomes is heterozygousfor each of the traits.

Explain whathomologous chromosomes are.

Interpret Scientific IllustrationsCan you identify homologous chromosomes? Homologouschromosomes are paired chromosomes having genes for thesame trait located at the same place on the chromosome. Thegene itself, however, may have different alleles, producing dif-ferent forms of the trait.

Solve the ProblemThe diagram below shows chromosome 1 with four differentgenes present. These genes are represented by the letters F, g,h, and J. Possible homologous chromosomes of chromosome 1are labeled 2–5. Examine the five chromosomes and the genesthey contain to determine which of chromosomes 2–5 arehomologous with chromosome 1.

Thinking Critically1. Classify Could chromosome 2 be homologous with

chromosome 1? Explain.2. Classify Could chromosome 3 be homologous with



chromosome 1? Explain.

1

K

m

n

O

4

F

g

J

h

5

F

g

J

h

2

F

G

j

h

3

F

G

K

h


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2 Teach

PurposeStudents will compare alleles onhomologous chromosomes.

Process Skillsinterpret scientific diagrams,apply concepts, compare and con-trast, think critically

Teaching StrategiesReview the meaning of alleles andhomologous chromosomes.

Thinking Critically1. 2 is homologous with 1. Alleles

may or may not be identical,as long as they are positionedat the same location onmatching chromosomes.

2. 3 is not homologous with 1.Genes must be identical.Gene K is not the same asgene J.

3. 4 is not homologous with 1.Chromosomes must match inphysical size and location ofgenes in order to be homolo-gous. (Sex chromosomes arean exception to the rule ofmatching physical size.)

4. 5 is homologous with 1 forthe same reasons as in question 1.

AssessmentPerformance Ask students tomake a diagram of a chromosomewith three marked gene locations.Have them diagram all thehomologous chromosomes thatare possible for this chromosome.Use the Performance TaskAssessment List for ScientificDrawing in PASC, p. 127. L2

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Everett Anderson Introduce studentsto the contribution of African Americancell biologist Everett Anderson to themodern understanding of the meioticprocess. Anderson (1928), who receivedhis Ph.D. in 1955, has been one of theleading researchers in developing

electron microscopic techniques tostudy meiosis. Obtain a copy ofAnderson’s 1972 publication, The MeioticProcess: Pairing, Recombination, andChromosome Movements, and discusswith students the methodologies used instudying the process of meiosis.

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Why meiosis?When cells divide by mitosis, the

new cells have exactly the same num-ber and kind of chromosomes as theoriginal cells. Imagine if mitosis werethe only means of cell division. Eachpea plant parent, which has 14 chro-mosomes, would produce gametesthat contained a complete set of 14 chromosomes. That means thateach offspring formed by fertilizationof gametes would have twice the num-ber of chromosomes as each of its par-ents. The F1 pea plants would havecell nuclei with 28 chromosomes, andthe F2 plants would have cell nucleiwith 56 chromosomes.

Clearly, there must be anotherform of cell division that allows off-spring to have the same number ofchromosomes as their parents. Thiskind of cell division, which producesgametes containing half the numberof chromosomes as a parent’s bodycell, is called meiosis (mi OH sus).Meiosis occurs in the specialized bodycells of each parent that producegametes.

Meiosis consists of two separate divi-sions, known as meiosis I and meiosisII. Meiosis I begins with one diploid(2n) cell. By the end of meiosis II, there are four haploid (n) cells. These

haploid cells are called sex cells—gametes. Male gametes are calledsperm. Female gametes are calledeggs. When a sperm fertilizes an egg,the resulting zygote once again hasthe diploid number of chromosomes.

10.2 MEIOSIS 265(tl)Rob & Ann Simpson/Visuals Unlimited, (tr)Aaron Haupt, (b)Robert J. Ellison/Photo Researchers

Chromosome 4

a

Terminal

Tall

Inflated

Axial

Short

Constricted

T

D

t

d

A

Homologous Chromosome 4

Figure 10.10Each chromosome 4 in garden peas contains genes for flower position,pod shape, and height, among others. Flower position can be either axial(flowers located along the stems) or terminal (flowers clustered at the topof the plant). Pod shape can be either inflated or constricted. Plant heightcan be either tall or short.

Table 10.1 Chromosome Numbers of Common Organisms

Organism Body Cell (2n) Gamete (n)

Fruit fly 8 4

Garden pea 14 7

Corn 20 10

Tomato 24 12

Leopard frog 26 13

Apple 34 17

Human 46 23

Chimpanzee 48 24

Dog 78 39

Adder’s tongue fern 1260 630

Adder’stongue fern

Corn

Leopardfrog

(p. 264)Homologous chromosomesare the pairs of chromo-somes in a diploid cell thatnormally contain identicalarrangements of genes.

Visual LearningTable 10.1 Ask students whetherthere is any evidence to supportthe idea that plants have fewerchromosomes than animals. Tell students to use examplesfrom the table to support theiranswers. No; apples have 34chromosomes, which is morethan fruit flies or frogs have.Ask for the chromosome numbersin skin cells of a leopard frog anda dog, 26 and 78 and in root cellsof tomatoes and garden peas. 24and 14

InquiryDiscussion Show students an eggand explain that it is a gamete.Based on previous information,students should be able to tellhow many alleles are present foreach trait. one Ask why an organ-ism cannot produce gametes bymitosis. Mitosis produces a cellwith two alleles for each trait.

Learning Disabled: Kinesthetic Havestudents use pipe cleaners or jelly beansto show why meiosis is necessary toprevent the number of chromosomesfrom doubling in each generation.Students should start with a cell con-taining two pipe cleaners or jelly beans,2n. After meiosis, gametes should each

contain one, n. After fertilization, thezygote has two. Students can repeatthe process for several generations,assuming gametes are formed by mito-sis. Students will see that the numberof chromosomes doubles in each gener-ation.

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The zygote then develops by mitosisinto a multicellular organism. This pat-tern of reproduction, involving theproduction and subsequent fusion ofhaploid sex cells, is called sexualreproduction. It is illustrated inFigure 10.11.

Explain why meiosisis necessary in organisms.

The Phases of MeiosisDuring meiosis, a spindle forms

and the cytoplasm divides in the sameways they do during mitosis.However, what happens to the chro-mosomes in meiosis is very different.Figure 10.12 illustrates interphaseand the phases of meiosis. Examinethe diagram and photo of each phaseas you read about it.

InterphaseRecall from Chapter 8 that, dur-

ing interphase, the cell replicates itschromosomes. The chromosomes are

replicated during interphase that pre-cedes meiosis I, also. After replication,each chromosome consists of two iden-tical sister chromatids, held together bya centromere.

Prophase I A cell entering prophase I behaves

in a similar way to one enteringprophase of mitosis. The DNA of thechromosomes coils up and a spindleforms. As the DNA coils, homologouschromosomes line up with each other,gene by gene along their length, toform a four-part structure called atetrad. A tetrad consists of two homol-ogous chromosomes, each made up oftwo sister chromatids. The chro-matids in a tetrad pair tightly. In fact,they pair so tightly that non-sisterchromatids from homologous chro-mosomes can actually break andexchange genetic material in a processknown as crossing over. Crossingover can occur at any location on achromosome, and it can occur at sev-eral locations at the same time.


Figure 10.11 In sexual reproduction, the dou-bling of the chromosome num-ber that results from fertilizationis balanced by the halving of thechromosome number that resultsfrom meiosis.

Fertilization

Meiosis

Mitosis andDevelopment

Diploid zygote(2n = 46)

Haploid gametes(n = 23)

Sperm cell

Multicellular diploid adults

(2n = 46)

Meiosis

Egg cell

meiosis from theGreek word meioun,meaning “to dimin-ish”; Meiosis is celldivision that resultsin a gamete contain-ing half the numberof chromosomes ofits parent.

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Modeling Meiosis

PurposeStudents will observe the changes thatoccur during meiosis.

Safety Precautions

Preparationcut 9 sheets of unlined paper or posterboard into 30-cm squaresMaterials long length of yarn, paper clips, longlength of string, toothpicks, tape or glue,scissors ProcedureGive students the following directions.1. Work in groups of nine students. Each

student is to model one phase of inter-phase and meiosis. Each model is to

have a phase name, labels, and anexplanation of the events taking place.

2. Represent cell structures as follows: yarnstrands � chromosomes, paper clips �

centromeres, string � nuclear mem-branes, toothpicks � spindle fibers.

3. Place models on large sheets of paper orposter board.

Phases of Meiosis Have eight students form two“tetrads” and walk throughthe phases of meiosis.

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Visual LearningFigure 10.12 Ask students howthe processes of mitosis and meio-sis are different. Events that occurin meiosis but not in mitosisinclude (1) in prophase I, pairs ofhomologous chromosomes formtetrads and crossing over canoccur; (2) in metaphase I, homol-ogous chromosomes line up inpairs rather than independently;(3) in anaphase I, homologouschromosomes, not chromatids,separate; (4) in telophase I, eachcell has only one chromosomefrom each pair; and (5) at theend of meiosis II, each new cellhas the haploid number of chromosomes.

Meiosis is nec-essary to provide variationand maintain the diploidnumber of chromosomesafter reproduction.

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10.2 MEIOSIS 267(clockwise from top) (1–4 7 9)John D. Cunningham/Visuals Unlimited, (5 6 8)Carolina Biological Supply/Phototake, NYC

Figure 10.12Compare these diagrams of meiosis with those of mitosis in Chapter 8. After telophase II,meiosis is finished and gametes form. Compare and Contrast In what other ways aremitosis and meiosis different?

Meiosis I

Metaphase I

Metaphase II

Prophase I

Prophase II

Interphase

Telophase II

Telophase IAnaphase II

Anaphase I

Meiosis II

LM Magnification: 255�




Stained LM Magnification: 580�Stained LM Magnification: 580�


Stained LM Magnification: 580�


Using an AnalogyTo reinforce the concept ofhomologous chromosomes, sisterchromatids, tetrad formation,crossing over, and anaphase I, trythe following analogy: A magicpair of shoes (left and right) isfound on a shelf (homologouschromosomes in a cell). Theseshoes, being magic, can and doreplicate (interphase replication).

Each copy is tied to its originalwith its shoelaces (centromere;both lefts are tied together andboth rights are tied together).Both rights are now called rightsister shoes (sister chromatids).Both lefts are now called left sistershoes (sister chromatids). All fourshoes line up next to one another(tetrad formation).

While next to one another,part of one nonsister shoe (a leftshoe) exchanges its innersole withanother nonsister shoe (a rightshoe) (crossing over). Right shoesmove away from left shoes to dif-ferent shelves (anaphase I, withhomologous chromosomes sepa-rating and going to two differentcells. Both rights are still tiedtogether and both lefts are stilltied together).

To improve the analogy, locatetwo identical pairs of shoes todemonstrate the events as theyare described.

Caption Question AnswerFigure 10.12 There are twodivisions in meiosis and only onein mitosis. Meiosis produceshaploid cells and mitosis pro-duces diploid cells.

4. Only one pair of chromosomes is to befollowed through all models. Glue ortape all parts in place.

5. Model interphase, prophase I,metaphase I, anaphase I, and telophase I,prophase II, metaphase II, anaphase II,and telophase II.

Expected ResultsStudents will gain an understanding ofmeiosis when visualizing each phase.

Analysis1. What happens to the chromosome

number during meiosis? reduced by 1/2 2. How many cells are formed during

meiosis? 43. What is the fate of the cells formed dur-

ing meiosis? They form either egg orsperm cells.

AssessmentKnowledge Ask students to write aparagraph explaining the value of mak-ing models such as the ones in this lab.Use the Performance Task AssessmentList for Writing in Science in PASC, p. 159. L2

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It is estimated that during prophase Iof meiosis in humans, there is an aver-age of two to three crossovers for eachpair of homologous chromosomes.

This exchange of genetic material is diagrammed in Figure 10.13B.Crossing over results in new combina-tions of alleles on a chromosome, asyou can see in Figure 10.13C. Youcan practice modeling crossing over inthe MiniLab at the left.

Metaphase IDuring metaphase I, the centromere

of each chromosome becomes attachedto a spindle fiber. The spindle fiberspull the tetrads into the middle, orequator, of the spindle. This is animportant step unique to meiosis.Note that homologous chromosomesare lined up side by side as tetrads. Inmitosis, on the other hand, they lineup on the spindle’s equator independ-ently of each other.

Anaphase I Anaphase I begins as homologous

chromosomes, each with its two chro-matids, separate and move to oppositeends of the cell. This separation occursbecause the centromeres holding thesister chromatids together do not splitas they do during anaphase in mitosis.This critical step ensures that each newcell will receive only one chromosomefrom each homologous pair.

Telophase IEvents occur in the reverse order

from the events of prophase I. Thespindle is broken down, the chromo-somes uncoil, and the cytoplasmdivides to yield two new cells. Each cellhas half the genetic information of theoriginal cell because it has only onechromosome from each homologouspair. However, another cell division isneeded because each chromosome isstill doubled.


Formulate ModelsModeling Crossing OverCrossing over occurs duringmeiosis and involves onlythe nonsister chromatidsthat are present duringtetrad formation. Theprocess is responsible forthe appearance of newcombinations of alleles ingamete cells.

Procedure! Copy the data table.@ Roll out four long strands of clay at least 10 cm long to

represent two chromosomes, each with two chromatids. # Use the figure above as a guide to joining and labeling

these model chromatids. Although there are four chro-matids, assume that they started out as a single pair ofhomologous chromosomes prior to replication. The figureshows tetrad formation during prophase I of meiosis.

$ First, assume that no crossing over takes place. Model theappearance of the chromosomes in the four gamete cellsthat will result at the end of meiosis. Record your model’sappearance by drawing the gametes’ chromosomes andtheir genes in your data table.

% Next, repeat steps 2–4. This time, however, assume thatcrossing over occurs between genes B and C.

Analysis1. Predict What will be the appearance of the chromosomes

prior to replication?2. Compare Are there any differences in the combinations

of alleles on chromosomes in gamete cells when crossingover occurs and when it does not occur?

3. Analogy Crossing over has been compared to “shufflingthe deck” in cards. Explain what this means.

4. Think Critically What would be accomplished if crossingover occurred between sister chromatids? Explain.

5. Evaluate Does your model adequately represent crossingover in a cell?

Data Table

No Crossing Over Crossing Over

Appearance of chromosomes Appearance of chromosomes

2 chromosomes with chromatids

Nonsister chromatids

Twist tie

Mark geneswith a pencilpoint.

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Purpose Students will model the process ofcrossing over.

Process Skillsapply concepts, compare and con-trast, formulate models, recognizecause and effect, think critically,define operationally

Teaching Strategies� Plasticene or clay is available

from biological supply housesor craft stores.

� String or twine may be used asa substitute for twist ties.

� Review the concept that gam-etes contain only one chromo-some from each pair.

� Make sure students wash theirhands after the lab.

Expected ResultsAll gametes will show the samecombination of genes if no cross-ing over occurs and differentcombinations of genes if crossingover does occur.

Analysis1. Only two chromosomes

should be drawn. Thesequence of genes would beC B A on one chromosomeand c B a on the other.

2. Yes, crossing over is theexchange of genetic materialbetween nonsister chro-matids during prophase I ofmeiosis.

3. Student answers will vary.There is a mixing of the genetraits from their original orderas would occur in shufflingcards.

4. There would be no mixing ofgene traits when comparedwith the original chromo-somes because sister chro-matids are identical.

5. Student answers will vary. Forthe purposes of explainingcrossing over, the model issimplistic but accurate.

Modified AssessmentPerformance Ask students to modelthe appearance of gamete cells if cross-ing over occurred between sister chro-matids. This should confirm theiranswer to question 5. Use thePerformance Task Assessment List forModel in PASC, p. 123.

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The phases of meiosis IIThe newly formed cells in some

organisms undergo a short restingstage. In other organisms, however,the cells go from late anaphase ofmeiosis I directly to metaphase ofmeiosis II.

The second division in meiosis issimply a mitotic division of the prod-ucts of meiosis I. Meiosis II consists ofprophase II, metaphase II, anaphase II,and telophase II. During prophase II, aspindle forms in each of the two newcells and the spindle fibers attach to thechromosomes. The chromosomes, stillmade up of sister chromatids, arepulled to the center of the cell and lineup randomly at the equator duringmetaphase II. Anaphase II begins as thecentromere of each chromosome splits,allowing the sister chromatids to sepa-rate and move to opposite poles.Finally, nuclei re-form, the spindlesbreak down, and the cytoplasm dividesduring telophase II. The events ofmeiosis II are identical to those youstudied for mitosis except that thechromosomes do not replicate beforethey divide at the centromeres.

At the end of meiosis II, four hap-loid cells have been formed from onediploid cell. Each haploid cell con-tains one chromosome from eachhomologous pair. These haploid cellswill become gametes, transmittingthe genes they contain to offspring.

Meiosis Provides forGenetic Variation

Cells that are formed by mitosis areidentical to each other and to the par-ent cell. Crossing over during meiosis,however, provides a way to rearrangeallele combinations. Rather than thealleles from each parent stayingtogether, new combinations of allelescan form. Thus, variability is increased.

Genetic recombinationHow many different kinds of sperm

can a pea plant produce? Each cellundergoing meiosis has seven pairs ofchromosomes. Because each of theseven pairs of chromosomes can lineup at the cell’s equator in two differ-ent ways, 128 different kinds of spermare possible (2n � 27 � 128).

10.2 MEIOSIS 269

pro- from theGreek word pro,meaning “before”meta- from theGreek word meta,meaning “after”ana- from theGreek word ana,meaning “up,back, again”telo- from theGreek telos,meaning “end”The four phases ofcell division areprophase,metaphase,anaphase, andtelophase.

Homologous chromosomesCrossing over in tetrad

Gametes

Tetrad

A A

Sister chromatids Nonsister chromatids

B B

a a

b b b

a a

B Bb b

b

AA

BB

a a

A AFigure 10.13 Late in prophase I, the homologouschromosomes come together to formtetrads (A). Arms of nonsister chro-matids wind around each other (B), andgenetic material may be exchanged (C).

AA BB

CC

Using an AnalogyContinue the shoe analogy ordemonstration of meiosis to illus-trate telophase I, metaphase II,and telophase II events.

Start with two right shoes tiedtogether. Both sets are separatefrom each other on different closetshelves (two new cells formed aftertelophase I). The shoes untie (cen-tromere splits after metaphase II).They move to different shelves inthe closet (two new cells formed intelophase II).

Ask students: (a) How manyshoes are now on separate shelves(separate cells)? 4 (b) How manyshoes were present at the start ofthis analogy? 2 (c) Is the chromo-some number in a cell diploid orhaploid before a cell undergoesmeiosis? diploid (d) How manyshoes are present on each shelf atthe end of the process? 1 (e) Howmany shoes were on the originalshelf? 2 (f) Is the chromosomenumber in each of the four newcells formed in meiosis reducedby half? yes

Concept DevelopmentThe chances of two humans beingborn exactly alike is almost animpossibility (except for identicaltwins). Explain why children inthe same family can neverthelessresemble one another ratherclosely.

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Meiosis Use models of plant or animal cell meiosis (available frombiological supply houses) to help students visualize the process.

Pages 268–269: Biology/Life Sciences 2a, 2d, 7dInv. & Exp. 1i


Figure 10.14 If a cell has two pairs ofchromosomes—A and a,B and b (n � 2)—fourkinds of gametes (22)are possible, dependingon how the homologouschromosomes line up atthe equator duringmeiosis I (A). This eventis a matter of chance.When zygotes areformed by the union ofthese gametes, 22 � 22

or 16 possible combina-tions may occur (B).

ab

aB

Ab

AB

AB

AABB AABb AaBB AaBb

AABb AAbb AaBb Aabb

AaBB AaBb aaBB aaBb

AaBb Aabb aaBb aabb

Possible combinations ofchromosomes in zygotes (in boxes)

Poss

ible

com

bina

tion

ofch

rom

osom

es in

egg

s

Possible combination ofchromosomes in sperm

Ab aB ab

Possible gametesPossible gametes

MEIOSIS I

MEIOSIS II

Chromosome A Chromosome aChromosome B Chromosome b

In the same way, any pea plant can form 128 different eggs. Becauseany egg can be fertilized by anysperm, the number of different possi-ble offspring is 16 384 (128 � 128). A simple example of how geneticrecombination occurs is shown inFigure 10.14A. You can see that thegene combinations in the gametesvary depending on how each pair of homologous chromosomes lines up during metaphase I, a randomprocess.

These numbers increase greatly asthe number of chromosomes in thespecies increases. In humans, n � 23,so the number of different kinds ofeggs or sperm a person can produce ismore than 8 million (223). When fertil-ization occurs, 223 � 223, or 70 trillion,different zygotes are possible! It’s nowonder that each individual is unique.

In addition, crossing over can occuralmost anywhere at random on a chro-mosome. This means that an almostendless number of different possiblechromosomes can be produced bycrossing over, providing additionalvariation to the variation already pro-duced by the random assortment ofchromosomes. This reassortment of

chromosomes and the genetic infor-mation they carry, either by crossingover or by independent segregation ofhomologous chromosomes, is calledgenetic recombination. It is a majorsource of variation among organisms.Variation is important to a speciesbecause it is the raw material thatforms the basis for evolution.

Explain how cross-ing over increases genetic variability.

Meiosis explains Mendel’s resultsThe behavior of the chromosomes

in meiosis provides the physical ba-sis for explaining Mendel’s results.The segregation of chromosomes inanaphase I of meiosis explains Men-del’s observation that each parentgives one allele for each trait at ran-dom to each offspring, regardless ofwhether the allele is expressed. Thesegregation of chromosomes at ran-dom during anaphase I also explainshow factors, or genes, for differenttraits are inherited independently ofeach other. Today, Mendel’s laws andthe events of meiosis together formthe foundation of the chromosometheory of heredity.

AA BB

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Crossing overchanges some of the alleleson a chromosome to thealternate alleles.

PortfolioComparing Mitosis and MeiosisProvide students with two sets ofsimple line drawings, one showinginterphase and the phases of mito-sis and the other showing inter-phase and meiosis. Place compara-ble phases next to each otherwhen possible (prophase of mito-sis next to prophase I of meiosis,and so on). Ask students todescribe in their portfolios thesimilarities and differencesbetween processes. Also havethem indicate the type of cellformed at the end of the process—body or gamete—and whether itschromosome number would bediploid or haploid. L2

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Modeling the Membrane Have stu-dent groups invent a mechanism fordemonstrating how the plasma mem-brane of an animal cell pinchestogether during telophase I and II of

meiosis. Students can use a variety ofcommon objects as models, such asballoons, string, and wire.

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NondisjunctionAlthough the events of meiosis usu-

ally proceed accurately, sometimeschromosomes fail to separate correctly.The failure of homologous chromo-somes to separate properly duringmeiosis is called nondisjunction.Recall that during meiosis I, onechromosome from each homologouspair moves to each pole of the cell. Innondisjunction, both chromosomesof a homologous pair move to thesame pole of the cell.

In one form of nondisjunction, twokinds of gametes result. One has anextra chromosome, and the other ismissing a chromosome. The effects ofnondisjunction are often seen aftergametes fuse. For example, when agamete with an extra chromosome isfertilized by a normal gamete, thezygote will have an extra chromo-some. This condition is called trisomy(TRI soh mee). In humans, if a gametewith an extra chromosome number 21is fertilized by a normal gamete, theresulting zygote has 47 chromosomesinstead of 46. This zygote will developinto a baby with Down syndrome.

Although organisms with extrachromosomes often survive, organ-isms lacking one or more chromo-somes usually do not. When a gametewith a missing chromosome fuseswith a normal gamete during fertil-ization, the resulting zygote lacks achromosome. This condition is calledmonosomy. In humans, most zygoteswith monosomy do not survive. If azygote with monosomy does survive,the resulting organism usually doesnot. An example of monosomy that isnot lethal is Turner syndrome, inwhich human females have only a single X chromosome instead of two.

Another form of nondisjunctioninvolves a total lack of separation ofhomologous chromosomes. When thishappens, a gamete inherits a completediploid set of chromosomes, like thoseshown in Figure 10.15. When agamete with an extra set of chromo-somes is fertilized by a normal haploidgamete, the offspring has three sets ofchromosomes and is triploid. Thefusion of two gametes, each with anextra set of chromosomes, producesoffspring with four sets of chromo-somes—a tetraploid.

Male parent (2n)

Meiosis

Abnormalgamete (2n)

Nondisjunction

Zygote(4n)

Female parent (2n)

Meiosis

Abnormalgamete (2n)

Nondisjunction

Figure 10.15 Follow the steps to seehow a tetraploid plant,such as this chrysanthe-mum, is produced.

R.B. Satterthwaite

DisplayMake a bulletin board display ofdisorders caused by abnormalnumbers of chromosomes, whichresult from nondisjunction.Include a diagram, similar toFigure 10.15, showing the steps ofnondisjunction. You may want toinclude photos, descriptions, andsymptoms of each disorder.

Some examples of nondisjunc-tion are:Trisomy 21—Down syndromeTrisomy 13—Patau’s syndromeXO—Turner’s syndromeXXX—Trisomy X (metafemales)XXY—Klinefelter’s syndromeXYY—Jacob’s syndromeOY—males (lethal)

Polyploidy in Plants: LinguisticHave students research and reporton how plant breeders create poly-ploidy flowers. Students shoulddraw some of the plants and indi-cate how many sets of chromosomeseach has. L3

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Chromosome MappingFigure 10.16Crossing over, the exchange of genetic material by nonsisterchromatids, provides information that can be used to make chro-mosome maps. Crossing over occurs more frequently betweengenes that are far apart on a chromosome than between genesthat are closer together. Critical Thinking Why is the fre-quency of crossing over related to the distance betweengenes on a chromosome?

272 MENDEL AND MEIOSISB. John/Cabisco/Visuals Unlimited

AA BB50

AA DD BB CC10 5

DD AA CC BB510

or or

CC DD

DD CC35

35

or

AA BBCCDD

50

10 35 5

Crossing over In prophase I ofmeiosis, nonsister chromatids crossover, as shown in the photo above.Each X-shaped region is a crossover.

AA

Mapping Crossing over produces new allele combinations.Geneticists use the frequency of crossing over to map the relativepositions of genes on a chromosome. Genes that are farther aparton a chromosome are more likely to have crossing over occurbetween them than are genes that are closer together.

BB

Frequencies and distance Suppose there are four genes—A, B, C, and D—ona chromosome. Geneticists determine that the frequencies of recombination amongthem are as follows: between A and B—50%; between A and D—10%; between Band C—5%; between C and D—35%. The recombination frequencies can beconverted to map units: A–B � 50; A–D � 10; B–C � 5; C–D � 35. These map unitsare not actual distances on the chromosome, but they give relative distancesbetween genes. Geneticists line up the genes as shown above.

CC

Making the mapThe genes can bearranged in thesequence that reflectsthe recombinationdata. This sequence is a chromosome map.

DD

Stained TEM Magnification: 1905�

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PurposeStudents study the process ofcrossing over and its importancein genetic recombination.

Teaching StrategiesAsk students to describe theprocess of crossing over.

Visual Learning� Have students examine the

photograph and count the chiasmata—regions where chro-mosomes appear to be physicallylinked.

� Visual-Spatial Have studentsdraw chromosomes before andafter crossing over. The use ofcolored pencils will help stu-dents visualize how this processleads to genetic recombination.

Critical ThinkingThe farther apart two alleles arefrom each other on a chromo-some, the greater the chancethat the chromosome will be tan-gled, broken, and rearrangedbetween them.

Caption Question AnswerFigure 10.17 There are noseeds in the fruits.

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Chromosome Mapping Have interestedstudents map the four genes on a chro-mosome given that the frequencies ofrecombination between the genes areas follows: A and D—60%, A and

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Understanding Main Ideas1. How are the cells at the end of meiosis different

from the cells at the beginning of meiosis? Usethe terms chromosome number, haploid, anddiploid in your answer.

2. What is the significance of meiosis to sexualreproduction?

3. Why are there so many varied phenotypes withina species such as humans?

4. If the diploid number of a plant is 10, how manychromosomes would you expect to find in itstriploid offspring?

Thinking Critically5. How do the events that take place during

meiosis explain Mendel’s law of independentassortment?

6. Get the Big Picture Compare Figures 10.12 and8.13 of meiosis and mitosis. Explain why crossingover between nonsister chromatids of homolo-gous chromosomes cannot occur during mitosis.For more help, refer to Get the Big Picture in theSkill Handbook.

SKILL REVIEWSKILL REVIEW

10.2 MEIOSIS 273Bildarchiv Okapi/Photo Researchers

PolyploidyOrganisms with more than the usual

number of chromosome sets are calledpolyploids. Polyploidy is rare in animalsand almost always causes death of thezygote. However, polyploidy frequentlyoccurs in plants. Often, the flowers andfruits of these plants are larger thannormal, and the plants are healthier.Many polyploid plants, such as the ster-ile banana plant shown in Figure 10.17,are of great commercial value.

Meiosis is a complex process, andthe results of an error occurring aresometimes unfortunate. However, theresulting changes can be beneficial,such as those that have occurred inagriculture. Hexaploid (6n) wheat,triploid (3n) apples, and polyploidchrysanthemums all are availablecommercially. You can see that a thor-ough understanding of meiosis andgenetics would be very helpful toplant breeders. In fact, plant breedershave learned to produce polyploidplants artificially by using chemicalsthat cause nondisjunction.

Gene Linkage and MapsGenes sometimes appear to be

inherited together instead of inde-pendently. If genes are close togetheron the same chromosome, they usually

are inherited together. These genesare said to be linked. In fact, all thegenes on a chromosome usually arelinked and inherited together. It is thechromosomes, rather than the individ-ual genes, that follow Mendel’s law ofindependent assortment.

Linked genes may become sepa-rated on different homologous chro-mosomes as a result of crossing over.When crossing over produces newgene combinations, geneticists canuse the frequencies of these new genecombinations to make a chromosomemap showing the relative locations ofthe genes. Figure 10.16 illustratesthis process.

Figure 10.17The banana plant is anexample of a triploidplant. Think CriticallyWhy do you think thebanana plant is sterile?

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1. The chromosome number in a cell atthe end is half the chromosome num-ber in a parent cell. The original cellhas a diploid number of chromosomesand each of the new cells has a hap-loid number.

2. Meiosis reduces the chromosome num-ber in gametes. When sexual repro-duction occurs, the zygote recombines

the chromosomes and maintains the2n condition.

3. Crossing over as well as the reassort-ment of the 46 chromosomes bothcontribute to the large number ofphenotypes that are possible.

4. 155. After meiosis, only one member of

each homologous chromosome pair

can be found in a gamete. Thus, nogamete will end up with two homo-logues. Alleles on different chromo-somes will sort independently fromone another.

6. Tetrad formation does not occur dur-ing mitosis. This prevents crossing overfrom taking place.

Check for UnderstandingAsk students to explain howthe words in the followingcombinations are related.

a. diploid—haploid b. homologous chromosomes

—allelec. sperm—egg—zygoted. meiosis—gamete e. crossing over—genetic

recombination

ReteachAsk students to prepare a listof important characteristics ofeach step in meiosis. Thenhave them prepare a secondlist of the reasons meiosis isimportant to organisms thatreproduce sexually.

ResourcesFor more practice, useReading Essentials for Biology,Section 10.2.

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ExtensionOogenesis and spermatogenesisare terms used to describe egg andsperm cell formation in humansthrough meiosis. Have studentsresearch how these processes differand how they are alike.

Skill Provide students with simpleline diagrams of various phases ofmeiosis. The diagrams should notbe in sequence. Have studentsplace the diagrams in propersequence, name each phase, anddescribe what is occurring. L1

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Before YouBegin

It’s difficult to predict thetraits of plants if all thatyou see is their seeds. Butif these seeds are plantedand allowed to grow, cer-tain traits will appear. Byobserving these traits, youmight be able to deter-mine the possible pheno-types and genotypes ofthe parent plants that pro-duced these seeds. In thislab, you will determinethe genotypes of plantsthat grow from twogroups of tobacco seeds.Each group of seeds camefrom different parents.Plants will be either greenor albino (white) in color.Use the following geno-types for this cross. CC � green, Cc � green,and cc � albino

How can phenotypes and genotypes of plants be determined?

ProblemCan the phenotypes and genotypes of the parent plants thatproduced two groups of seeds be determined from the phe-notypes of the plants grown from the seeds?

HypothesesHave your group agree on a hypothesis to be tested that willanswer the problem question. Record your hypothesis.

ObjectivesIn this BioLab, you will:� Analyze the results of growing two groups of seeds.� Draw conclusions about phenotypes and genotypes based

on those results.� Use the Internet to collect and compare data from other

students.

Possible Materialspotting soil light sourcesmall flowerpots or seedling thermometer or

flats temperature probetwo groups of tobacco seeds plant-watering bottlehand lens

Safety PrecautionsCAUTION: Always wash your hands after handling plantmaterials. Always wear goggles in the lab.

Skill HandbookIf you need help with this lab, refer to the Skill Handbook.

1. Examine the materials provided by your teacher. As agroup, make a list of the possible ways you might testyour hypothesis.

2. Agree on one way that your group could investigate yourhypothesis.

PLAN THE EXPERIMENTPLAN THE EXPERIMENT

PREPARATIONPREPARATION

Matt Meadows


274

PROCEDUREPROCEDURE

Troubleshooting� Students should keep growing conditions

for the two seed groups as constant as pos-sible. Soil should be kept moist at all times.Natural window light should be sufficient.Temperature should be kept above 20ºC.

� Seeds should be planted about 1 cm belowthe soil. Planting is easier if the soil is moist.

� Be sure students mark the seed type plantedin the flat or pot. Craft sticks can serve asmarkers.

Time Allotmentinitial session: one class period;follow-up session: 5 minutes eachday for watering, 20 minutes onlast day for counting

Process Skillshypothesize, observe and infer,collect data

Safety PrecautionsSome seed materials are poison-ous. Do not allow students to eatthe seeds. Have students washtheir hands after the lab.

Alternative MaterialsSeeds can be germinated andobserved in petri dishes. This willeliminate the need for soil, flats,or pots. Place seeds on moistenedpaper towels in the bottom of thedish and keep the dish covered.

Teaching Strategies� When supplying seeds to stu-

dents, make sure that the twotypes are kept separate fromeach other. Stick seeds onto apiece of tape for dispensing.

� An ideal quantity of seeds touse is 20–30 per type.

� Cotyledons (seed leaves) willappear after about 10 days.

Possible Hypotheses� If the parent plants were true

breeding for green color, thenall offspring will be green.

� If the parent plants were het-erozygous for green color, thenoffspring will show an approxi-mate ratio of 3 green to 1 white.

PREPARATIONPREPARATION

Pages 274–275: Biology/Life Sciences 2g, 3, 3aInv. & Exp. 1a, 1j

10.2 MEIOSIS 275

ANALYZE AND CONCLUDEANALYZE AND CONCLUDE

1. Think Critically Why was it necessary to grow plants from the seeds inorder to determine the phenotypes of the plants that formed the seeds?

2. Draw Conclusions Using the information in the introduction, describehow the gene for green color (C) is inherited.

3. Make Inferences For the group of seeds that yielded all green plants, areyou able to determine exactly the genotypes of the parents that formed theseseeds? Can you determine the genotype of each plant observed? Explain.

4. Make Inferences For the group of seedsthat yielded some green and some albinoplants, are you able to determine exactlythe genotypes of the plants that formedthese seeds? Can you determine the geno-type of each plant observed? Explain.

5. Use the data posted onto

compare your experimental design withthat of other students. Were your resultssimilar? What might account for the differences?

ERROR ANALYSIS

Find this BioLab using the link below andpost your results in the table provided.Briefly describe your experimental design.

3. Design an experiment that will allow you to collect quanti-tative data. For example, how many plants do you thinkyou will need to examine?

4. Prepare a numbered list of directions. Include a listof materials and the quantities you will need.

5. Make a data table for recording your observations.

Check the Plan1. Carefully determine what data you are going

to collect. How many seeds will you need? Howlong will you carry out the experiment?

2. What variables, if any, will have to be con-trolled? (Hint: Think about the growing condi-tions for the plants.)

3. Make sure your teacher has approved your experi-mental plan before you proceed further.

4. Carry out your experiment. Make any needed obser-vations, such as the numbers of green and albino plantsin each group, and complete your data table.

5. Visit to post your data.6. Make wise choices in the disposal of

materials.CLEANUP AND DISPOSAL

Ray Pfortner/Peter Arnold, Inc.

ca.bdol.glencoe.com/internet_lab

ca.bdol.glencoe.com/internet_lab ca.bdol.glencoe.com/internet_lab

1. Leaf color cannot beobserved in the seed butappears only after the planthas emerged from the seed.

2. The gene for green color is adominant trait.

3. No, one parent may havebeen true breeding for green(CC), the other may havebeen heterozygous (Cc). Thiswould have yielded all greenoffspring. Offspring may beCC or Cc but still appeargreen.

4. Yes, both parents must beheterozygous to yield a ratioof 3 green to 1 albino. Forthe offspring genotypes, youcan conclude only that thealbino offspring are cc.Green are either CC or Cc.

5. Answers may vary. Geneticratios are governed by thelaws of probability. The larg-er the population size, thecloser the calculated valuewill be to the theoretical.

AssessmentPortfolio Ask students to makediagrams that show the parentaland offspring genotypes and phe-notypes for both groups of seedsused in this experiment. Use thePerformance Task AssessmentList for Scientific Drawing inPASC, p. 127. L2

P

LS

ANALYZE AND CONCLUDEANALYZE AND CONCLUDE

275

Data and ObservationsSeeds that came from true breeding plantswill produce plants that are all green. Seedsfrom heterozygous parents will produce bothgreen and albino seedlings in the ratio ofabout 3 green to 1 albino. Have studentsreview Problem-Solving Lab 10.1 for help incalculating the ratio of green to albino plants.

To navigate the InternetBioLabs, go to

. Studentsshould visit this site only afterthey have done the experimentand collected their data to posttheir experimental design andcompare it with that of otherstudents. Students should becareful that only data fromidentical crosses are pooled.

ca.bdol.glencoe.com/internet_lab

http://ca.bdol.glencoe.com/internet_lab




Students may say that the ratios revealedthat a dominant trait showed up threetimes more often because it was alwaysable to overcome the effect of a reces-sive trait that accompanied it.

276

PurposeStudents will gain insight into theimportance of mathematics to thestudy of biology. They will learnhow mathematics aided Mendelin understanding the laws ofheredity.

Teaching Strategies

� Students may be surprised tolearn that a biologist of CharlesDarwin’s stature missed a greatopportunity in his brilliantcareer. Had he interpreted hisdata about snapdragon flowershapes correctly, the wholeworld might have understoodthe laws of heredity 40 yearsearlier.

� Discuss how ratios helpedMendel see that definite factorswere being passed on from par-ents to offspring. He didn’tknow what these factors were,nor how they operated. Heknew nothing about chromo-somes and meiosis, yet was ableto show how traits were trans-mitted because of his mathe-matical analysis.


A Solution from Ratios

In 1866, Gregor Mendel, an Austrian monk,published the results of eight years of experi-

ments with garden peas. His work was ignoreduntil 1900, when it was rediscovered.

Mendel had three qualities that led to his dis-covery of the laws of heredity. First, he was curi-ous, impelled to find out why things happened.Second, he was a keen observer. Third, he was askilled mathematician. Mendel was the first biol-ogist who relied heavily on statistics for solu-tions to how traits are inherited.

Darwin missed his chance About the sametime that Mendel was carrying out his experi-ments with pea plants, Charles Darwin was gath-ering data on snapdragon flowers. When Darwincrossed plants that had normal-shaped flowerswith plants that had odd-shaped flowers, all theoffspring had normal-shaped flowers. He thoughtthe two traits had blended. When he allowed theF1 plants to self-pollinate, his results were 88plants with normal-shaped flowers and 37 plantswith odd-shaped flowers. Darwin was puzzled bythe results and did not continue his studies withthese plants. Lacking Mendel’s statistical skills,Darwin failed to see the significance of the ratioof normal-shaped flowers to odd-shaped flowersin the F2 generation. What was this ratio? Was itsimilar to Mendel’s ratio of dominant to recessivetraits in pea plants?

Finding the ratios for four other traitsFigure 10.3 on page 256 shows seven traits thatMendel studied in pea plants. You have alreadylooked at Mendel’s data for plant height andseed shape. Now use the data for seed color,flower position, pod color, and pod shape to findthe ratios of dominant to recessive for thesetraits in the F2 generation.

Draw Table B in your notebook or journal.Calculate the ratios for the data in Table A andcomplete Table B by following these steps:

• Step 1 Divide the larger number by the smaller number.

• Step 2 Round to the nearest hundredth.• Step 3 To express your answer as a ratio,

write the number from step 2 followed by a colon and the number 1.

R.W. VanNorman/Visuals Unlimited

Think Critically Why are ratios so important inunderstanding how dominant and recessive traitsare inherited?

To find out more about Mendel’s work, visit ca.bdol.glencoe.com/math

Table A Mendel’s Results

Seed Flower Pod PodColor Position Color Shape

Yellow Axial Green Inflated6022 651 428 882

Green Terminal Yellow Constricted2001 207 152 299

Table B Calculating Ratios for Mendel’s Results

Seed Flower Pod PodColor Position Color Shape

Calculation 6022� 3.00

2001

Ratio 3:1yellow: green

651� 3.14 428

� 2.82 882� 2.95

207 152 299

Pages 276–277: Biology/Life Sciences 2a, 2d, 2e, 3,3bInv. & Exp. 1n

http://ca.bdol.glencoe.com/math

Section 10.1STUDY GUIDESTUDY GUIDE

CHAPTER 10 ASSESSMENT 277

Section 10.2

To help you reviewMendel’s work, use the OrganizationalStudy Fold on page 253.

Key Concepts� Genes are located on chromosomes and

exist in alternative forms called alleles. Adominant allele can mask the expression ofa recessive allele.

� When Mendel crossed pea plants differingin one trait, one form of the trait disap-peared until the second generation of off-spring. To explain his results, Mendelformulated the law of segregation.

� Mendel formulated the law of independentassortment to explain that two traits areinherited independently.

� Events in genetics are governed by thelaws of probability.

Vocabularyallele (p. 256)dominant (p. 256)fertilization (p. 253)gamete (p. 253)genetics (p. 253)genotype (p. 258)heredity (p. 253)heterozygous (p. 259)homozygous (p. 258)hybrid (p. 255)law of independent

assortment (p. 260)law of segregation

(p. 257)phenotype (p. 258)pollination (p. 254)recessive (p. 256)trait (p. 253)zygote (p. 253)

Mendel’s Lawsof Heredity

Key Concepts� In meiosis, one diploid (2n) cell produces

four haploid (n) cells, providing a way foroffspring to have the same number ofchromosomes as their parents.

� In prophase I of meiosis, homologous chro-mosomes come together and pair tightly.Exchange of genetic material, called cross-ing over, takes place.

� Mendel’s results can be explained by the dis-tribution of chromosomes during meiosis.

� Random assortment and crossing over dur-ing meiosis provide for genetic variationamong the members of a species.

� The outcome of meiosis may vary due tonondisjunction, the failure of chromosomesto separate properly during cell division.

� All the genes on a chromosome are linkedand are inherited together. It is the chro-mosomes rather than the individual genesthat are assorted independently.

Vocabularycrossing over (p. 266)diploid (p. 263)egg (p. 265)genetic recombination

(p. 270)haploid (p. 263)homologous chromo-

some (p. 264)meiosis (p. 265)nondisjunction (p. 271)sexual reproduction

(p. 266)sperm (p. 265)

Meiosis

ca.bdol.glencoe.com/vocabulary_puzzlemaker

277

Key ConceptsSummary statements can beused by students to review themajor concepts of the chapter.

For additional helpwith vocabulary,have students

access the VocabularyPuzzleMaker online at

Visit /self_check_quiz/vocabulary_puzzlemaker/chapter_test/standardized_test

Use the ExamView®Pro Test Bank CD-ROM to:• Create multiple versions of tests• Create modified tests with one mouse click for inclusion students• Edit existing questions and add your own questions• Build tests aligned with state standards using built-in

State Curriculum Tags• Change English tests to Spanish with one mouse click and vice versa

FOLDABLES™Have students use theirFoldables to review the contentof Section 10.1. On the back ofthe paper, use Mendel’s laws ofheredity to determine the off-spring from the following cross:AaBbCc � AaBbCc.

ca.bdol.glencoe.com

ca.bdol.glencoe.com/vocabulary_puzzlemaker


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Review the Chapter 10 vocabulary words listed inthe Study Guide on page 277. For each set ofvocabulary words, choose the one that does notbelong. Explain why it does not belong.

1. egg—sperm—zygote2. homozygous—hybrid—heterozygous3. phenotype—genotype—allele4. nondisjunction—genetic recombination—

crossing over5. zygote—diploid—gamete

6. At the end of meiosis, how many haploidcells have been formed from the originalcell?A. one C. threeB. two D. four

7. When Mendel transferred pollen from onepea plant to another, he was ________ theplants.A. self-pollinating C. self-fertilizingB. cross-pollinating D. cross-fertilizing

8. Which of these does NOT show a recessivetrait in garden peas?A. B. C. D.

9. During what phase of meiosis do sister chromatids separate?A. prophase I C. anaphase IIB. telophase I D. telophase II

10. During what phase of meiosis do nonsisterchromatids cross over?A. prophase I C. telophase IB. anaphase I D. telophase II

11. A dihybrid cross between two heterozygotesproduces a phenotypic ratio of ________.A. 3:1 C. 9:3:3:1B. 1:2:1 D. 1:6:9

12. Open Ended On the average, each humanhas about six recessive alleles that would belethal if expressed. Why do you think thathuman cultures have laws against marriagebetween close relatives?

13. Open Ended How does separation of homol-ogous chromosomes during anaphase I ofmeiosis increase variation among offspring?

14. Open Ended Relating to the methods ofscience, why do you think it was importantfor Mendel to study only one trait at a timeduring his experiments?

15. Open Ended Explain why sexual reproduc-tion is an advantage to a population thatlives in a rapidly changing environment.

16. Observe and Infer Why is it possible tohave a family of six girls and no boys, butextremely unlikely that there will be a publicschool with 500 girls and no boys?

17. Recognize Cause and Effect Why is itsometimes impossible to determine the genotype of an organism that has a domi-nant phenotype?

18. Observe and Infer While examining a cellin prophase I of meiosis, you observe a pairof homologous chromosomes pairing tightly.What is the significance of the places atwhich the chromosomes are joined?

19. Several humangenetic disorders result from nondisjunctionin meiosis, including Down syndrome,Kleinfelter’s syndrome, and Turner syn-drome. Visit to inves-tigate these disorders. What characteristic iscommon to each? Choose one of these dis-orders, or another human disorder causedby nondisjunction, and prepare a visual dis-play that explains the disorder. Explain thedisorder to your class.

REAL WORLD BIOCHALLENGE

278 CHAPTER 10 ASSESSMENT

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Questions andanswers have been

aligned and verified by ThePrinceton Review.

1. A zygote results from theunion of egg and sperm.

2. Homozygous contains twoidentical alleles.

3. Alleles are the factors thatproduce genotypes andphenotypes

4. Nondisjunction is a failure ofcrossing over and its result-ing genetic recombination.

5. Gametes are haploid.

6. D 9. C7. B 10. A8. D 11. C

12. The likelihood that closerelatives share the samerecessive genes is greaterthan in the general popula-tion, thus raising the risk ofa child being homozygousfor those traits.

13. The order of lining up at theequator during metaphase Iof meiosis will vary, thus pro-viding additional variationwhen the chromosomes sep-arate during anaphase I.

14. Controlled experiments,such as Mendel’s, requirethat no more than one vari-able be manipulated at atime. By doing many experi-ments, Mendel was able todetermine the principlesthat govern genetics. Theseprinciples would not havebeen easily observable with-out controlled experiments.

15. Sexual reproduction pro-vides for variations, some ofwhich will be able to surviveand reproduce duringchanging conditions.

278

16. The probability of a family with six girlsis (1/2)6, but the probability of anentire school of girls would be (1/2)500.

17. If the dominant allele completely masksthe recessive allele, you cannot tell if anorganism with the dominant trait ishomozygous or heterozygous for thedominant trait.

18. These are places where crossing overtakes place and genetic material isexchanged.

19. All three disorders result from nondis-junction. Students may make displaysthat show karyotypes or examples ofpeople with the disorders.


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Multiple ChoiceUse the diagram to answer questions 20–23.

20. Which of the following is true?A. Individual 1 is heterozygous.B. Individuals 2 and 3 are homozygous.C. Individual 4 is recessive.D. All individuals will be male.

21. Which of the following has the Tt genotype?A. 1 C. 3B. 2 D. 2 and 3

22. If T is the allele for purple flowers and t isthe allele for white flowers, the resultswould be ________.A. 3 out of 4 are purpleB. 3 out of 4 are whiteC. equal numbers of white and purpleD. all of the same color

23. Which of Mendel’s observations woulddescribe the results of the experimental cross in question 22?A. rule of dominanceB. law of segregationC. law of independent assortmentD. rule of unit factors

24. Recessive traits appear only when an organism is ________.A. matureB. different from its parentsC. heterozygousD. homozygous

25. The stage of meiosis shown here is ________.A. anaphase IB. metaphase IIC. telophase ID. telophase II

Study the diagram and answer questions 26–28.

26. What name is given to the process shownabove?A. fertilization C. meiosisB. zygote D. gametes

27. What name is given to the cells shown inthe diagram above?A. fertilization C. meiosisB. zygotes D. gametes

28. If each of the cells shown in the diagram has16 chromosomes, how many chromosomeswould you expect to find in a skin cell of theresulting organism?A. 16 C. 32B. 64 D. 8

T

T 1 2

43t

t

CHAPTER 10 ASSESSMENT 279

Constructed Response/Grid InRecord your answers on your answer document.

29. Open Ended Explain the difference between trisomy and triploidy. Describe a way that eachcondition could occur. Use diagrams to clarify your answer.

30. Open Ended Compare metaphase of mitosis with metaphase I of meiosis. Explain the significance ofthe differences between the two stages in terms of sexual reproduction and genetic variation.

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3a

3a

3a

3a

2a

2e3b

The assessed California standard appears next to the question.

CHAPTER ASSESSMENT CHAPTER 10 Mendel and Meiosis 35

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AssessmentStudent Recording Sheet

Name Date Class

Chapter

10Chapter Chapter AssessmentChapter Assessment

Use with pages 278–279of the Student Edition

Standardized Test Practice

Vocabulary ReviewWrite the word you chose and explain why it does not belong.

1. _________________________________________________________________________________

2. _________________________________________________________________________________

3. _________________________________________________________________________________

4. _________________________________________________________________________________

5. _________________________________________________________________________________

Understanding Key ConceptsSelect the best answer from the choices given and fill in the corresponding oval.

6. 9.

7. 10.

8. 11.

Constructed ResponseRecord your answers for Questions 12–15 on a separate sheet of paper.

Thinking CriticallyRecord your answers for Questions 16–18 on a separate sheet of paper.

19. Follow your teacher’s instructions for presenting your BioChallenge answer.REAL WORLD BIOCHALLENGE

Part 2Constructed Response/Grid InRecord your answers for Questions 29 and 30 on a separate sheet of paper.

A B C D A B C D

A B C D A B C D

A B C DA B C D

Part 1 Multiple ChoiceSelect the best answer from the choices given andfill in the corresponding oval.

20. 25.

21. 26.

22. 27.

23. 28.

24.

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D

A B C D

279

20. C 23. B 26. A21. D 24. D 27. D22. A 25. C 28. C29. Trisomy—three chromo-

somes of the same type.Triploidy—three sets ofchromosomes in a cell.

30. In mitosis, metaphase is thestage where duplicatedchromosomes line up on theequator. Each sister chro-matid is attached to a spin-dle fiber, which extends toopposite poles. The sisterchromatids will separateinto two new cells. There isno genetic variation in thesenew cells. In meiosis, meta-phase I is the point wherealleles on homologous chro-mosomes pair up on theequator, and spindle fibersare attached to each chro-mosome on opposing poles.The new cells formed arehaploid. If the alleles are dif-ferent, then half of thedaughter cells will carry agenetic variation differentfrom the other.

For more help evaluatingopen-ended assessmentquestions, see the rubric on p. 9T.

California StandardsPractice

Pages 278–279: Biology/Life Sciences 2a, 2d, 2e, 2g,3, 3a

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unit overview genetics - temecula valley unified …web1.tvusd.k12.ca.us/gohs/myoung/bio....

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