Section 1 The Origins of Genetics - Jourdanton ISD · Before beginning this section review with your students the objectives listed in the Student ... TEKS Bio 3F, 6A, 6D, 10A Teacher

OverviewBefore beginning this sectionreview with your students theobjectives listed in the StudentEdition. This section explainsMendel’s discoveries in modernterms and explains traits expressedas ratios.

Ask students to list on paper fivecharacteristics that are passed on in families (eye, hair and skin color,height, and so on), and to name one characteristic that may also beinherited but that is also influencedby behavior or environment (musclesize, body weight, having a sun tan,and so on). Intrapersonal

Demonstration Display large pictures of a few flower-ing plants or bring in real plants.Ask students to come up with a listof traits that could be inherited inplants. Encourage students to thinkof many different traits, such asflower shape, flower color, flowerposition on stem, leaf shape, leafcolor, pattern of veins, pattern ofstem growth, presence of hairs onstems, and inner structure of flower.Ask them if they think the traits areinherited together or separately.

Visual TAKS 2 Bio 6DLS

MotivateMotivate

LS

Bellringer

FocusFocus

Section 1

162 Chapter 8 • Mendel and Heredity

• Lesson Plan• Directed Reading• Active Reading• Data Sheet for Math Lab GENERAL

GENERAL

GENERAL

Chapter Resource File• Reading Organizers• Reading Strategies• Problem Solving Worksheet

Ratios and Proportions GENERAL

Planner CD-ROM

Transparencies

TT BellringerTT Three Steps of Mendel’s

Experiment

Objectives● Identify the investigator

whose studies formed thebasis of modern genetics.

● List characteristics that makethe garden pea a good sub-ject for genetic study.

● Summarize the three majorsteps of Gregor Mendel’sgarden-pea experiments.

● Relate the ratios that Mendelobserved in his crosses tohis data.

Key Terms

hereditygeneticsmonohybrid crosstrue-breedingP generationF1 generationF2 generation

Section 1 The Origins of Genetics

Mendel’s Studies of TraitsMany of your traits, including the color and shape of your eyes, thetexture of your hair, and even your height and weight, resemblethose of your parents. The passing of traits from parents to offspringis called . Humans have long been interested in heredity.From the beginning of recorded history, we have attempted to altercrop plants and domestic animals to give them traits that are moreuseful to us. Before DNA and chromosomes were discovered, hered-ity was one of the greatest mysteries of science.

Mendel’s Breeding ExperimentsThe scientific study of heredity began more than a century ago withthe work of an Austrian monk named Gregor Johann Mendel, shownin Figure 1. Mendel carried out experiments in which he bred different varieties of the garden pea Pisum sativum, shown in Figure 2and in Table 1. British farmers had performed similar breedingexperiments more than 200 years earlier. But Mendel was the first todevelop rules that accurately predict patterns of heredity. The pat-terns that Mendel discovered form the basis of , the branchof biology that focuses on heredity.

Mendel’s parents were peasants, so he learned much about agri-culture. This knowledge became invaluable later in his life. As a young man, Mendel studied theology and was ordained as apriest. Three years after being ordained, he went to the University ofVienna to study science and mathematics. There he learned how tostudy science through experimentation and how to use mathematicsto explain natural phenomena.

Mendel later repeated the experi-ments of a British farmer, T. A. Knight.Knight had crossed a variety of the gar-den pea that had purple flowers with avariety that had white flowers. (Theterm cross refers to the mating orbreeding of two individuals.) All of theoffspring of Knight’s crosses had pur-ple flowers. However, when two of thepurple-flowered offspring were crossed,their offspring showed both white andpurple flowers. The white trait hadreappeared in the second generation!

Mendel’s experiments differed fromKnight’s because Mendel counted thenumber of each kind of offspring andanalyzed the data.

genetics

heredity

Figure 1 Gregor Mendel.Mendel’s experiments with garden peas led to our modern understanding of heredity.

3F

6D

3F

3F

162

TAKS 2

Student Edition TAKS Obj 2 Bio 6A TAKS Obj 2 Bio 6D TAKS Obj 2 Bio 10A TEKS Bio 3F, 6A, 6D, 10A

Teacher Edition TAKS Obj 1 Bio/IPC 3C TAKS Obj 2 Bio 6A, 6DTEKS Bio 3B, 6A, 6DTEKS Bio/IPC 3C

pp. 162–163

Trends in GenomicsCats and Humans Researchers, working onthe genomes of organisms have found that whenit comes to the arrangement of genes on ourchromosomes, we are closer to cats than to anyother groups studied so far except for primates.Stephen J. O’Brien, a geneticist and chief of theNation Cancer Institute’s laboratory of GenomicDiversity, began studying the genetics of thehouse cat in the 1970’s. The Cat Genome Projectis a comprehensive genetic analysis of Feliscatus. The results of this research have proveduseful in boosting human AIDS research andhave been useful in criminal forensics. Bio/IPC 3C

K-W-L Before they read this chap-ter, have each student write a shortlist of all the things they alreadyKnow (or think they know) aboutinheritance. Ask them to contributetheir entries to a group list on theboard or overhead projector. Thenhave the students list things theyWant to know about inheritance.Have students save their lists forlater used in section 4.

Teaching TipGenetic make-up Bring photos or stuffed toys of animals with different traits. Use these props to emphasize that many genes areinvolved in giving an animal itsoverall appearance, and that thegenes for most traits have two ormore versions. Ask them is theycan estimate how many genes animals have in common with eachother. For example, chimpanzeesand humans share approximately98% of their genetic makeup.

Visual

Group Activity Benefits of Peas Divide the classinto small groups. Have each groupdesign newspaper ads that wouldhave attracted someone like Mendelto purchase peas for geneticresearch. The ads should mentionall of the benefits of Pisum sativumthat make it useful for geneticresearch. Ask students to use illustrations in their ads. Encouragestudents to be creative. They mayuse butcher paper, computer, con-struction paper, and so on. Post theads on the bulletin board and leada discussion on the benefits of thegarden-pea for genetic research.Bio 3B

TAKS 2 Bio 6A, 6D (grade 10 only)LS

GENERALSKILLBUILDER

READINGREADING

TeachTeach

Chapter 8 • Mendel and Heredity 163

BIOLOGYBIOLOGY• Unit 5—Heredity: Introduction

This engaging tutorial introduces students to principles and practicalapplications of Mendelian genetics.

English Language Learners

Table 1 The Seven Traits Mendel Studied and Their Contrasting FormsFlower color

Seed shape

Pod color

Pod shape

Flower position

Plant height

Seed color

Useful Features in PeasThe garden pea is a good subject for studying heredity for severalreasons:

1. Several traits of the garden pea exist in two clearly differentforms. For example, the flower color is either purple or white—there are no intermediate forms. Table 1 shows the seven traitsthat Mendel chose to study.

2. The male and female reproductive parts of garden peas areenclosed within the same flower. You can control mating byallowing a flower to fertilize itself (self-fertilization), or you cantransfer the pollen to another flower on a different plant (cross-pollination). To cross-pollinate two pea plants, Mendel removedthe stamens (the male reproductive organs that produce pollen)from the flower of one plant. As shown in Figure 2, he thendusted the pistil (the female reproductive organ that produceseggs) of that plant with pollen from a different pea plant.

3. The garden pea is small, grows easily, matures quickly, and pro-duces many offspring. Thus, results can be obtained quickly, andthere are plenty of subjects to count.

Mendel transferred pollen froma second flower to the pistil ofthe original flower.

To cross-pollinate flowers ofdifferent colors, Mendel firstremoved the stamens—thepollen-producing structures—from one flower.

Figure 2 Pollen transfer in Mendel’s experiments

163


Trends in GeneticsFlies and Worms Many scientists who study genetics use the fruit fly Drosophilamelanogaster or the roundworm Caenorhabditisin their research. These organisms show a variety of traits, are easy to obtain and breed,have short generation time (less than 2 weeksfor fruit flies; less than 3 days for roundworms),and produce a large number of offspring. Howlong would it take to study three generations of humans? TAKS 2 Bio 6D (grade 10 only)

Traits Expressed as Simple RatiosMendel’s initial experiments were monohybrid crosses. A

is a cross that involves one pair of contrastingtraits. For example, crossing a plant with purple flowers and a plantwith white flowers is a monohybrid cross. Mendel carried out hisexperiments in three steps, as summarized in Figure 3.

Step Mendel allowed each variety of garden pea to self-pollinatefor several generations. This ensured that each variety was

for a particular trait; that is, all the offspringwould display only one form of the trait. For example, atrue-breeding purple-flowering plant should produce onlyplants with purple flowers in subsequent generations.

These true-breeding plants served as the parental gener-ation in Mendel’s experiments. The parental generation, or

, are the first two individuals that are crossedin a breeding experiment.

Step Mendel then cross-pollinated two P generation plants thathad contrasting forms of a trait, such as purple flowers and white flowers. Mendel called the offspring of the P generation the first filial generation, or . Hethen examined each F1 plant and recorded the number ofF1 plants expressing each trait.

Step Finally, Mendel allowed the F1 generation to self-pollinate.He called the offspring of the F1 generation plants the sec-ond filial generation, or . Again, each F2 plantwas characterized and counted.

F2 generation

F1 generation

P generation

true-breeding

monohybrid cross

Figure 3

The word filial is from theLatin filialis, meaning “of ason or daughter.” Thus F (filial) generations are allthose generations that follow a P (parental) generation.

164

Student Edition TAKS Obj 1 Bio/IPC 2C TAKS Obj 2 Bio 6D TAKS Obj 3 Bio 13ATEKS Bio 3F, 6D, 13ATEKS Bio/IPC 2C

Teacher Edition TAKS Obj 1 Bio/IPC 2CTAKS Obj 2 Bio 6D TEKS Bio 5A, 6DTEKS Bio/IPC 2C

pp. 164–165

Teaching TipHidden traits Ask students if theycan tell by looking at the purplepea flowers in Figure 3 which onesare true-breeding for the purpletrait and which ones are not. Pointout that you cannot always tell thegenetic makeup of an organism bylooking at it. Ask students how across helps determine if a plant istrue-breeding for a trait.

Using the Figure Point out to students the male andfemale flower structures illustratedin Figure 2. Explain the differencebetween cross-pollination and self-pollination, and the significance ofremoving the stamens from theflower on the left. (In cross-pollina-tion, the pollen from one flower istransported to the female structuresof a different flower. In self-pollina-tion, the pollen from one flower istransported to the female structuresof the same flower. By removing themale stamens of a flower, it cannotself pollinate and the genetic make-upof both parents can be determinedwith certainty.)

Math Skills Ask students to prac-tice reducing ratios to their simplestforms. Survey the class for somenumbers to work with. For example,ask how many students own a cat.Have them divide each number (classsize; cat owners) by the smallest num-ber (cat owners) and write it as a ratio.If there are 30 students in class and10 own a cat, the ratio is 30 to 10.Simplified, �

3100� � 3 and �

1100� � 1, the ratio

is 3:1. Logical TAKS 1 Bio/IPC 2CLS

GENERALBUILDERSKILL

Bio 5A

GENERAL

TAKS 2 Bio 6D(grade 10 only)

GENERAL

Teach, continuedTeach, continued

OverviewBefore beginning this sectionreview with your students theobjectives listed in the StudentEdition. This section explainsMendel’s discoveries in modernterms and explains the law of segregation and the law of inde-pendent assortment.

Tell students that a gardener noticedthat some of the flowers on herplants were white. In previous years,the flowers had been purple. Askstudents to write down their pro-posed explanation for this difference.Tell them they will be finding outmore on this topic as they read thissection. (Instead of buying hybridseeds from the store, she decided toplant pea seeds from the crop sheharvested the previous year. Her plantswere the F2 generation, which showsa 3:1 ratio of purple to white flowers).

IdentifyingPreconceptions Ask students if it is possible for anoffspring to have traits differentfrom both of their parents. Somestudents will respond that this isnot possible. Explain that somehidden traits in the parents cancombine together and appear in the offspring. An example wouldbe two right-handed parents having a left-handed child. TAKS 2 Bio 6D

(grade 10 only)

MotivateMotivate

TAKS 2 Bio 6D (grade 10 only)

Bellringer

FocusFocus

Section 2


CulturalAwarenessCulturalAwareness

Blood and Inheritance The Greek philoso-pher Aristotle associated inheritance withblood. He thought the blood carried heredi-tary information from the body’s variousstructures to the reproductive organs. Weknow this is not true, but the idea is ingrainedin many languages. For example, “blue

blood,” “blood stock,” and “It is in theblood” (English); “Corre en la sangre”(Spanish); “Bon sang ne peut mentir” and“celle est dans le sang” (French); and “Es liegtim Blute” and “von gutem Blut” (German) allassociate inheritance with blood. Bio 3F

Objectives● Describe the four

major hypotheses Mendeldeveloped.

● Define the terms homozy-gous, heterozygous, genotype,and phenotype.

● Compare Mendel’s two lawsof heredity.

Key Terms

alleledominantrecessivehomozygousheterozygousgenotypephenotypelaw of segregationlaw of independent

assortment

Section # A Head 1-line

A Theory of HeredityBefore Mendel’s experiments, many people thought offspring were ablend of the characteristics of their parents. For example, if a tallplant were crossed with a short plant, the offspring would be mediumin height. Mendel’s results did not support the blending hypothesis.Mendel correctly concluded that each pea has two separate “heritablefactors” for each trait—one from each parent. As shown in Figure 4,when gametes (sperm and egg cells) form, each receives only one ofthe organism’s two factors for each trait. When gametes fuse duringfertilization, the offspring has two factors for each trait, one fromeach parent. Today these factors are called genes.

Mendel’s HypothesesThe four hypotheses Mendel developed were based directly on theresults of his experiments. These four hypotheses now make up theMendelian theory of heredity—the foundation of genetics.

1. For each inherited trait, an individual has two copies ofthe gene—one from each parent.

2. There are alternative versions of genes. For example, thegene for flower color in peas can exist in a “purple” version

Section 2 Mendel’s Theory

Y = Gene forYyellow seeds

y = Gene forygreen seeds

Parent Parent

Gametes

Meiosis

Fertilization

1. During gameteformation (meio-sis), the twogenes separate.

2. During fertilization,each offspringreceives one versionof each gene (allele)from each parent.

Each parent has two separate “factors,”or genes, for a particular trait.

Figure 4 Mendel’s factors

3F

3F

6D

166

TAKS 2

Student Edition TAKS Obj 2 Bio 4B TAKS Obj 2 Bio 6A TAKS Obj 2 Bio 6D TAKS Obj 2 Bio 10A TEKS Bio 3F, 4B, 6A, 6D, 10A

Teacher Edition TAKS Obj 1 Bio/IPC 2C TAKS Obj 2 Bio 4B, 6D TEKS Bio 3F, 4B, 6D, 6E, TEKS Bio/IPC 2C

pp. 166–167

Using the Figure Have students look at Figure 5.Point out to the students that thestages of meiosis result in gametesthat have only one version of eachgene. During Meiosis 1, a cellcompletes two successive divisionsthat produce 4 cells, each with achromosome number that has been reduced by half.

Teaching TipDominant and Recessive Ask students why some traits appearmore often than others do. Forexample, there are more dark hairedpeople than light haired people. To emphasize the point, do a handcount of some contrasting traitssuch as eye color, tongue curlingand free ear lobes. The majority of students will be dominant for agiven trait. Discuss dominant andrecessive and explain that for thedominant trait to appear, only oneallele for the trait is needed, but forthe recessive trait to appear, bothalleles for the trait must be inher-ited. However, emphasize thatdominant phenotypes are notalways more common than reces-sive phenotypes. If there are veryfew alleles for a dominant pheno-type in a population, it will notoccur often.

Teaching TipGenotype and Phenotype Havestudents practice using the boldfaceterms in this section by providingseveral examples. For example, tellthem that the gene for plant heighthas two versions: T � tall and t � dwarf. Ask students to identifythe two alleles for plant height. (T and t). Write Tt, tt, and TT onthe board and ask students to iden-tify the genotype and phenotype of each set of alleles (genotypes—Tt, tt, TT; phenotypes: tall, dwarf,and tall). Ask students to identifywhether a plant with TT alleles is homozygous or heterozygous(homozygous). TAKS 1 Bio/IPC 2C,

TAKS 2 Bio 6D (grade10 only)


GENERAL

TAKS 2 Bio 4B; Bio 6E

TeachTeach


• Lesson Plan• Directed Reading• Active Reading• Data Sheet for Quick Lab GENERAL

GENERAL

GENERAL

Chapter Resource File• Reading Organizers• Reading Strategies• Portfolio Project

Genetics Project GENERAL

Planner CD-ROM

Transparencies

TT BellringerTT Mendel’s Factors

or a “white” version. Today the different versions of a geneare called its . As shown in Figure 4, an individualreceives one allele from each parent. Each allele can bepassed on when the individual reproduces.

3. When two different alleles occur together, one of them maybe completely expressed, while the other may have noobservable effect on the organism’s appearance. Mendeldescribed the expressed form of the trait as .The trait that was not expressed when the dominantform of the trait was present was described as .For every pair of contrasting forms of a trait thatMendel studied, the allele for one form of the trait wasalways dominant and the allele for the other form of thetrait was always recessive. For example, if a plant hasboth purple and white alleles for flower color butblooms purple flowers, then purple is the dominantform of the trait; white is the recessive form. This isshown in Figure 5.

4. When gametes are formed, the alleles for each gene in anindividual separate independently of one another. Thus,gametes carry only one allele for each inherited trait.When gametes unite during fertilization, each gametecontributes one allele. As shown in Figure 4, each parentcan contribute only one of the alleles because of the waygametes are produced during the process of meiosis.

Mendel’s Findings in Modern TermsGeneticists have developed specific terms and ways of representingan individual’s genetic makeup. For example, letters are often usedto represent alleles. Dominant alleles are indicated by writing thefirst letter of the trait as a capital letter. For instance, in pea plants,purple flower color is a dominant trait and is written as P. Recessivealleles are also indicated by writing the first letter of the dominanttrait, but the letter is lowercase. For example, white flower color isrecessive and is written as p.

If the two alleles of a particular gene present in an individual arethe same, the individual is said to be (hoh moh ZIEguhs) for that trait. For example, a plant with two white flower allelesis homozygous for flower color, as shown in Figure 5. The allele foryellow peas, Y, is dominant to the allele for green peas, y. A plantwith two yellow-pea alleles, YY, is homozygous for seed color.

If the alleles of a particular gene present in an individual are dif-ferent, the individual is (heht uhr oh ZIE guhs) forthat trait. As shown in Figure 5, a plant with one “purple flower”allele and one “white flower” allele is heterozygous for flower color.A plant with one “yellow pea” allele and one “green pea” allele isheterozygous for seed color.

heterozygous

homozygous

recessive

dominant

alleles

Figure 5 Recessive alleles.Alleles can be present but notexpressed. The allele for purpleflowers, P, is dominant to therecessive allele, p.

ppWhite flowers,homozygous

recessive

PPPurple flowers,homozygous

dominantPp

Purple flowers,heterozygous

167

Teaching TipGene Expression Point out thatthe environment may influence theexpression of some phenotypes (suchas freckles). Ask students for otherexamples (muscle size).

Activity Graphic Organizer Have studentswork in pairs to make a graphicorganizer to demonstrate the law of independent assortment (see bot-tom of this page). Ask students toillustrate their graphic organizerwith at least one example showingthe inheritance of two pairs of con-trasting traits. Have them write abrief explanation. Ask student tovolunteer to put their examples on the board or overhead projector.TAKS 1 Bio/IPC 2C, TAKS 2 Bio 6D (grade 10 only)

GENERAL

TAKS 2 Bio 6D(grade 10 only)

GENERAL


Identifying Dominant or Recessive Traits Skills AcquiredSummarizing,calculating, applyinginformation

Teacher’s Notes Emphasize that dominant phenotypes are not more com-mon than recessive phenotypes.Point out that the expression of some phenotypes (such asfreckles) may be influenced by the environment.

Analysis Answers 1. Answers will vary. 2. Answers will vary. 3. The recessive traits. Recessive

traits must be homozygous tobe expressed.

English Language Learners

Use this graphic organizer withActivity on this page.

Graphic Organizer

SsBb

SB sB Sb sb

Independent assortment

Identifying Dominant or Recessive TraitsYou can determine some of the genotypes and all of the pheno-types for human traits that are inherited as simple dominant orrecessive traits.

Materials

pencil, paper

Procedure

1. Make a table like the one atright. For each trait, circle thephenotype that best matchesyour own phenotype.

2. Determine how many students in your class shareyour phenotype by recordingyour results in a table on thechalkboard.

Analysis

1. Summarize the classresults for each trait.

2. Calculate the classdominant:recessive ratio foreach trait.

3. Critical ThinkingApplying Information Forwhich phenotypes in the tablecan you determine a person’sgenotype without ever havingseen his or her parents?Explain.

Dominant trait Recessive trait

Cleft chin No cleft

Dimples No dimples

Hair above knuckles Hairless fingers

Freckles No freckles

In heterozygous individuals, only the dominantallele is expressed; the recessive allele is presentbut unexpressed. An example of a human traitthat is expressed in a heterozygous individual isfreckles. Freckles F, is a dominant allele. Therecessive allele is f, no freckles. The recessiveallele may be present but not expressed. As shownin Figure 6, people who are heterozygous forfreckles (Ff ) will have freckles even though theyalso have the allele for no freckles, f.

The set of alleles that an individual has iscalled its (JEE noh tiep). The physi-cal appearance of a trait is called a (FEE noh tiep). Phenotype is determined bywhich alleles are present. For example, if Pp isthe genotype of a pea plant, its phenotype ispurple flowers. If pp is the genotype of a peaplant, its phenotype is white flowers. Whenconsidering seed color, if Yy is the genotype ofa pea plant, its phenotype is yellow seeds. If yyis the genotype of a pea plant, its phenotype isgreen seeds. Note that by convention, the dom-inant form of the trait is written first, followedby the lowercase letter for the recessive form ofthe trait.

phenotypegenotype

Figure 6 Dominent alleles. In het-erozygous individuals, freckles, F, is thedominant allele. Similarly, the allele for acleft chin is dominant to the allele for achin without a cleft.

6A 6D

168

TAKS 2 Bio 6A, 6D

Student Edition TAKS Obj 2 Bio 4B, 10A TAKS Obj 2 Bio 6A TAKS Obj 2 Bio 6D TEKS Bio 3A, 3F, 4B, 6A, 6D, 10A

Teacher Edition TAKS Obj 1 Bio/IPC 2C, 3A TAKS Obj 2 Bio 4B, 6A, 6D TEKS Bio 3A, 3F, 4B, 6DTEKS Bio/IPC 2C, 3A

pp. 168–169

TAKS 2

Answers to Section Review

1. Genes are pieces of DNA that code for a particular trait. There are alternative versions,or alleles, for each gene.

2. The Pp plant is heterozygous dominant; the PPplant is homozygous dominant.

3. The Bb rabbit has a black coat and the bbrabbit has a brown coat.

4. Bb � heterozygous; bb � homozygous

5. A weakness in the law of independent assort-ment is that it applies only to genes that arelocated on different chromosomes or that are




TAKS 2 Bio 6A

far apart on the same chromosome.

6. A. Incorrect. A pea plant that hastwo recessive traits would be called homozy-gous. B. Incorrect. A pea plant that has twodominant traits would be called homozygous. C. Correct. A pea plant that has one dominantand one recessive would be called heterozygous.D. Incorrect. Multiple alleles, genes with threeor more alleles, would describe a pea plant thathas three dominant and one recessive alleles.TAKS 2 Bio 6D (grade 10 only)

TAKS 1 Bio/IPC 3A; Bio 3F

Activity Hairy Knuckles Have studentsdetermine whether or not they havehair above their knuckles. Tell stu-dents that the presence of hairabove the knuckles is caused by adominant allele, H. Then ask themto identify the genotype of a personwho does not have hair above theirknuckles (hh). Have students deter-mine under what circumstances aparent without hair above theirknuckles can produce a child withhair above their knuckles. (The sec-ond parent must have the hair abovethe knuckles gene) Intrapersonal

ReteachingHave students review the stepsinvolved in Mendel’s scientificinvestigations. Then have themapply these methods to Mendel’sdiscoveries about heredity. Ask students to summarize Mendel’shypothesis and predictions.

Quiz1. How are the genotype of a domi-

nant allele and a recessive allelewritten? (A dominant allele is rep-resented by a capital letter, and thesame letter in lower case representsa recessive allele.)

2.What is the phenotype of a purple-flowered pea plant? (PP or Pp)

3. What is a dihybrid cross? (A cross that considers two pairsof contrasting traits.)

AlternativeAssessmentAsk students to relate Mendel’sfour hypotheses to his experimentalresults. Refer the students to Figure 3 for Mendel’s experimentalresults. TAKS 1 IPC 3A; Bio 3A

GENERAL

GENERAL

TAKS 1 IPC 3A; Bio 3A

CloseClose

TAKS 2 Bio 6A, 6D (grade 10 only)LS

GENERAL


Section 2 Review

Differentiate between alleles and genes. 6A

Apply the terms homozygous, heterozygous,dominant, or recessive to describe plants with the genotypes PP and Pp. 6D

Identify the phenotypes of rabbits with thegenotypes Bb and bb, where B = black coat and b = brown coat. 6D

Determine whether the rabbits in item 3 are heterozygous or homozygous. 6D

Critical Thinking Critiquing ExplanationsReview Mendel’s two laws according to theirstrengths and weaknesses in terms of our mod-ern understanding of meiosis. 3A 3F

If a pea plant is heterozygousfor a particular trait, how can the alleles that controlthe trait be characterized? 6D

A two recessive C one dominant, one recessive B two dominant D three dominant, one recessive

TAKS Test PrepTAKS Test Prep

The Laws of Heredity Mendel’s hypotheses brilliantly predicted the results of his crossesand also accounted for the ratios he observed. Similar patterns ofheredity have since been observed in countless other organisms.Because of their importance, Mendel’s ideas are often referred to asthe laws of heredity.

The Law of SegregationThe first law of heredity describes the behavior of chromosomesduring meiosis. At this time, homologous chromosomes and thenchromatids are separated. The first law, the ,states that the two alleles for a trait segregate (separate) whengametes are formed (as shown in Figure 4).

The Law of Independent AssortmentMendel went on to study whether the inheritance of one trait (suchas plant height) influenced the inheritance of a different trait (suchas flower color). To study how different pairs of genes are inherited,Mendel conducted dihybrid crosses. A dihybrid cross is a cross thatconsiders two pairs of contrasting traits. For example, a cross thatconsiders both plant height and flower color is a dihybrid cross.

Mendel found that for the traits he studied, the inheritance of onetrait did not influence the inheritance of any other trait. The

states that the alleles of different genesseparate independently of one another during gamete formation. Forexample, the alleles for the height of the plant shown in Figure 7 sep-arate independently of the alleles for its flower color. We now knowthat this law applies only to genes that are located on different chro-mosomes or that are far apart on the same chromosome.

The search for the physical nature of Mendel’s “factors” domi-nated biology for more than half a century after Mendel’s work wasrediscovered in 1900. We now know that the units of heredity areportions of DNA called genes, which are found on the chromosomesthat an individual inherits from its parents.

of independent assortmentlaw

law of segregation

Figure 7 The law of independent assortment.Mendel found that the inheritance of one trait, suchas plant height, did not influence the inheritance ofanother trait, such as flowercolor.

169

OverviewBefore beginning this sectionreview with your students theobjectives listed in the StudentEdition. This section explains the use of Punnett squares for predicting outcomes, probabilityand pedigrees.

Since the dawn of agriculture, peoplehave used selective breeding toimprove crops and domestic animals.Modern applications of Mendeliangenetics and gene technology haveresulted in major changes in cropsand animals. Ask students to list on paper some examples of selec-tive breeding in domestic animalsor crops. Ask students to explainhow they might go about selectingfor a particular trait.

Discussion/ Question Tell students that the basenji is adog that cannot bark. However,they can make a yodeling type ofsound. Basenjis are small dogs withpointed ears, short silky hair androws of wrinkles on their foreheads.Ask students to hypothesize agenetic explanation for why thebasenjis cannot bark. (The ability tobark is a dominant trait in dogs. Allbasenjis have two recessive genes forthis trait.) Ask them if they can sug-gest other traits that have beenselected for in dogs or cats.TAKS 2 Bio 6A, 6D (grade 10 only)

MotivateMotivate

Bio/IPC 3C

Bellringer

FocusFocus

Section 3


did you know?Point out to students that breeders use Punnettsquares to help them select individuals that willbe most likely to produce offspring of the phe-notype they want.

Objectives● Predict the results of mono-

hybrid genetic crosses byusing Punnett squares.

● Apply a test cross to deter-mine the genotype of anorganism with a dominantphenotype.

● Predict the results of mono-hybrid genetic crosses byusing probabilities.

● Analyze a simple pedigree.

Key Terms

Punnett squaretest crossprobabilitypedigreesex-linked trait

Section # A Head 1-lineSection 3 Studying Heredity

Punnett Squares Animal breeders try to breed animals with very specific character-istics. Thus, breeders must be able to predict how often a trait willappear when two animals are crossed (bred). Likewise, horticultur-ists (plant breeders) need to produce plants with very specificcharacteristics. One simple way of predicting the expected results(not necessarily the actual results) of the genotypes or phenotypesin a cross is to use a Punnett square.

A is a diagram that predicts the outcome of agenetic cross by considering all possible combinations of gametesin the cross. Named for its inventor, Reginald Punnett, the sim-plest Punnett square consists of four boxes inside a square. Asshown in Figure 8, the possible gametes that one parent can pro-duce are written along the top of the square. The possible gametesthat the other parent can produce are written along the left side ofthe square. Each box inside the square is filled in with two lettersobtained by combining the allele along the top of the box with theallele along the side of the box. The letters in the boxes indicatethe possible genotypes of the offspring.

One Pair of Contrasting Traits Punnett squares can be used to predict the outcome of a monohybridcross (a cross that considers one pair of contrasting traits betweentwo individuals). For example, a Punnett square can be used to pre-dict the outcome of a cross between a pea plant that is homozygousfor yellow seed color (YY) and a pea plant that is homozygous forgreen seed color (yy). Figure 8 shows that 100 percent of the off-spring in this type of cross are expected to be heterozygous (Yy),expressing the dominant trait of yellow seed color.

Punnett square

Yy

Yy

Y

y

yy

Y

Yy

Yy

yy(Homozygous recessive)

Possible gametesfrom each parent

4_4

= Yy (Heterozygous)

YY(Homozygous dominant)

A cross between a peaplant that is homozygousfor yellow seeds (YY ) anda pea plant that ishomozygous for greenseeds (yy ) will produceonly yellow heterozygousoffspring (Yy ).

Figure 8 Monohybrid cross: homozygous plants

2C 6D

6D

2C

2C 6D

170

TAKS 1, TAKS 2

TAKS 1, TAKS 2

TAKS 2

TAKS 1

Student Edition TAKS Obj 1 Bio/IPC 2C TAKS Obj 2 Bio 6A TAKS Obj 2 Bio 6D TEKS Bio 6A, 6DTEKS Bio/IPC 2C

Teacher Edition TAKS Obj 1 Bio/IPC 2C, 3CTAKS Obj 2 Bio 6A, 6D TEKS Bio 6A, 6DTEKS Bio/IPC 2C, 3C

pp. 170–171

Using the Figure Point out to students that Figure 8shows a Punnett square used forpredicting the outcome of a geneticcross. The genotype of a parentdetermines the possible alleles thatcan be found in their gametes. The possible gametes are writtenalong the top and left sides of thesquare. Review with students howthe genotype in each square wasobtained. Assign several monohybridcrosses for students to practice.

Teaching TipPossible gametes Have studentspractice listing possible gameteswhen given the genotype of a dihybrid parent. LogicalTAKS 1 Bio/IPC 2C

LS

TAKS 1 Bio/IPC 2C

TeachTeach

• Lesson Plan• Directed Reading• Active Reading• Data Sheet for Data Lab• Data Sheet for Math Lab GENERAL

GENERAL

GENERAL

GENERAL

Chapter Resource File• Reading Organizers• Reading Strategies• Problem Solving Worksheet

Genetics and Probability GENERAL

Planner CD-ROM

Crosses that Involve Two TraitsTeaching Strategies• Explain how genotypes are

written in dyhibrid crosses.(for example, RrYy, not RYry)

Discussion• Under what circumstances

might a dihybrid cross fail toproduce four different kindsof gametes? (if the genes areclose together on the samechromosome)

Answers• Possible genotypes—RRYY,

RRYy, RrYY, RrYy, Rryy,RRyy, rrYY, rrYy, rryy

• Genotypic ratio—1 RRYY, 2 RRYy, 1 RRyy, 2 RrYY, 4 RrYy, 2 Rryy, 1 rrYY, 2 rrYy, 1 rryy

• Possible phenotypes—roundyellow, round green, wrinkledyellow, wrinkled green

• Phenotypic ratio—9 round, yellow : 3 round, green : 3 wrinkled, yellow : 1 wrinkled,green

Transparencies

TT BellringerTT Monohybrid Crosses of Homozygous

Plants and of Heterozygous PlantsTT Probability with Two Coins


Figure 9 shows a Punnett square that predictsthe results of a monohybrid cross between twopea plants that are both heterozygous (Yy) forseed color. One-fourth of the offspring would beexpected to have the genotype YY, two-fourths (orone-half) would be expected to have the genotypeYy, and one-fourth would be expected to have thegenotype yy. Another way to express this is to saythat the genotypic ratio is 1 YY : 2 Yy : 1 yy.Because the Y allele is dominant over the y allele,three-fourths of the offspring would be yellow,and one-fourth would be green. The phenotypicratio is 3 yellow : 1 green.

Punnett squares allow direct and simple predic-tions to be made about the outcomes of geneticcrosses. Although animal breeders and horticul-turists are not always certain what characteristicswill turn up in the offspring, they can use the pre-dictions from Punnett squares to cross individualsthat they know will be most likely to produce off-spring with the desired phenotypes.

YY

yy

Y

Y

y

y

Yy(Heterozygous)

Yy

Yy

Yy(Heterozygous)

1_4 = yy (Homozygous recessive)

1_4 = YY (Homozygous dominant)

2_4 = Yy (Heterozygous)

Crosses That Involve Two Traits

Suppose a horticulturist has two traits that shewants to consider when crossing two plants. Across that involves two pairs of contrasting traitsis called a dihybrid cross. For example, she maywant to predict the results of a cross between twopea plants that are heterozygous for seed shape(R � round, r � wrinkled) and seed color (Y � yellow, y � green).

Determine possible gametesTo use a Punnett square to predict the results ofthis cross, first consider how the four alleles fromeither parent (RrYy) can combine to form gametesthat are either RY, Ry, rY, or ry (Figure A).

Then write these gametes on the top and left sidesof a Punnett square (Figure B).

Complete the Punnett squareOn a separate sheet of paper, make a copy of thePunnett square in Figure B, which has been par-tially filled in with the predicted genotypes. Fill inthe remaining genotypes, then do the following:• List all of the possible genotypes that can

result.• Calculate the genotypic ratio for this cross.• List all of the possible phenotypes that can

result.• Calculate the phenotypic ratio for this cross.

FurtherExploring Further

Parent

R r Y y(Round, yellow)

Possible gametes

Figure A Gametes

r yr YR yRY

RRYY RRYy RrYY RrYy

RRYy

RrYy

RrYY

RrYy

RY Ry rY ry

RY

Ry

rY

ry

YellowPossible

gametes from each parent

Figure B Punnett square

RrYy

RrYy

Yellow

Crossing two pea plantsthat are heterozygousfor seed color (Yy) willproduce offspring in the ratio shown in thePunnett square.

Figure 9 Monohybrid cross: heterozygous plants

2C 6A 6D

171

TAKS 1, TAKS 2

Teaching TipTest Cross Tell students that theyhave been presented with one ofMendel’s purple-flowering peaplants. Ask them if they can iden-tify the genotype of the plant inregard to flower color. Then havethem propose a method for discov-ering the genotype of the plant. Givethe students a clue by asking themwhat genotype can be determinedfrom the phenotype (homozygousrecessive). (The purple flowered pea plant should be crossed with awhite flowered pea plant; if any off-spring are white, the unknown washeterozygous)TAKS 1 Bio/IPC 2A, TAKS 2 Bio 6A

GENERAL



Analyzing a Test CrossSkills AcquiredAnalyzing, interpreting,inferring, drawing conclusions, predictingoutcomes

Teacher’s NotesEncourage students to recognizethe importance of sample size in making conclusions aboutthe genotype of an unknownindividual.

Answers to Analysis 1. possible alleles each parent can

produce2. the genotype of each possible

kind of offspring3. The genotypic ratio will be

4 Pp : 0 PP : 0 pp. The pheno-typic ratio for the offspringwill be 4 purple : 0 white.

4. heterozygous

TAKS 1 Bio/IPC 2C,TAKS 2 Bio 6D(grade 10 only)

010001011001110101000100100010011100100100010000010100100111010101001000101010010010

Determining Unknown Genotypes Animal breeders, horticulturists, and others involved in breedingorganisms often need to know whether an organism with a dominantphenotype is heterozygous or homozygous for a trait. How do theydetermine this? For example, how might a horticulturist determinewhether a pea plant with a dominant phenotype, such as yellowseeds, is homozygous (YY) or heterozygous (Yy)? The horticulturistcould perform a test cross. In a , an individual whosephenotype is dominant, but whose genotype is not known, is crossedwith a homozygous recessive individual.

For example, a plant with yellow seeds but of unknown genotype(Y?) is test-crossed with a plant with green seeds (yy). If all of the off-spring produce yellow seeds, the offspring must be Yy. Thus, thegenotype of the “unknown” plant must be YY. If half of the offspringproduce yellow seeds and half produce green seeds, the genotype ofthe unknown plant must be Yy. In reality, if the cross produces evenone plant that produces green seeds, the genotype of the unknownparent plant is likely to be heterozygous. After performing a testcross, the horticulturist can continue breeding the original plantwith more certainty of its genotype.

test cross

010001011001110101000100100010011100100100010000010100100111010101001000101010010010

Analysis

1. Determine what the lettersat the top and side of eachbox represent.

2. Determine what the lettersin each box represent.

3. Calculate the genotypicand phenotypic ratios thatwould be predicted if theparent of the unknown genotype were homozygous for the trait (Figure B).

4. Critical ThinkingPredicting OutcomesIf half of the offspring havewhite flowers, what is thegenotype of the plant withpurple flowers?

P

p

p

p P

p

p

P

Figure A Heterozygous (Pp) plant Figure B Homozygous (PP) plant

Is this purpleflowering peaplant Pp or PP?

Analyzing a Test Cross Background

You can use a test cross to determine whether a plant with pur-ple flowers is heterozygous (Pp) or homozygous dominant(PP). On a separate sheet of paper, copy the two Punnettsquares shown below, and fill in the boxes in each square.

2C 6D

www.scilinks.orgTopic: Breeding Texas

LivestockKeyword: HXX4003

172

TAKS 1, TAKS 2

Student Edition TAKS Obj 1 Bio/IPC 2C TAKS Obj 2 Bio 6D TEKS Bio 6DTEKS Bio/IPC 2C

Teacher Edition TAKS Obj 1 Bio/IPC 2A, 2C, 2D TAKS Obj 2 Bio 6A, 6D TEKS Bio 6A, 6DTEKS Bio/IPC 2A, 2C, 2D

pp. 172–173

IPC Benchmark Mini-Lesson

Biology/IPC Skills TAKS 1 Bio/IPC 2COrganize, analyze, and evaluate data.Activity Have students choose an imaginary reces-sive trait, such as furry feet. Ask students to design atest cross for determining the genotype of a personwithout furry feet.

Demonstration Shuffle a deck of cards. Ask stu-dents to determine the probabilityof drawing an ace from the deck.(Students may suggest �

542� or �

1130�.) Ask

how they arrived at this conclusion.Deal 13 cards from the top of thedeck. Count the number of aces inthose 13 cards and compare thatnumber with the students’ predic-tion. If the number varies from theprediction, have the students speculate about the reasons for the difference. Logical

Teaching TipUsing Probabilities in GeneticCrosses Point out that the probabil-ity of a specific genotype occurringin a cross can be obtained by settingup a Punnett square similar to thosein Figures 8 and 9. The probabilityof finding a specific allele in a gameteis written next to the possible alleleacross the top and along the side.TAKS 1 Bio/IPC 2C

GENERAL

TAKS 1 Bio/IPC 2C, 2DLS

GENERAL


Reviewing InformationBecause probability is aratio of a subset of all possi-ble outcomes to all possibleoutcomes, the value forprobability is never greaterthan 1. When it is less thanone, it can be expressed asa fraction or as a percent-age of the whole.

Outcomes of Crosses Like Punnett squares, probability calculations can be used to predictthe results of genetic crosses. is the likelihood that a spe-cific event will occur. Probabilities can be expressed in words, asdecimals, as percentages, or as fractions. For example, if an eventdefinitely will occur, its probability can be expressed as either 1 outof 1 (in words), 1 (as a decimal numeral), 100 percent (as a percent-age), or �

11

� (as a fraction). If an event definitely will not occur, itsprobability can be expressed as either 0 out of 0, 0, 0 percent, or �

00

�. In order to simplify our discussion of probability, we will

express probabilities as fractions. Probability can be determinedby the following formula:

Probability �

Consider the possibility that a coin tossed into the air will landon heads (one possible outcome). The total number of all possibleoutcomes is two—heads or tails. Thus, the probability that a coinwill land on heads is �

12

�, as shown in Figure 10.

Probability of a Specific Allele in a GameteThe same formula can be used to predict the probability of an allelebeing present in a gamete. If a pea plant has two alleles for seed color,the plant can contribute either allele (yellow or green) to the gameteit produces (the law of independent assortment). For a plant withtwo alleles for seed color, the total number of possible outcomes istwo—green or yellow. The probability that a gamete will carry theallele for green seed color is �

12

�. The probability that a gamete fromthis plant will carry the allele for yellow seed color is also �

12

�.

Probability of the Outcome of a CrossBecause two parents are involved in a genetic cross, both parentsmust be considered when calculating the probability of the out-come of a genetic cross. Consider the analogy of two coins beingtossed at the same time. The probability of a penny landing onheads is �

12

�, and the probability of a nickel landing on heads is �12

�. Theway one coin falls does not depend on how the other coin falls. Sim-ilarly, the allele carried by the gamete from the first parent does notdepend on the allele carried by the gamete from the second parent.The outcomes are independent of each other.

To find the probability that a combination of two independentevents will occur, multiply the separate probabilities of the twoevents. Thus, the probability that a nickel and a penny will bothland on heads is

�12

� � �12

� � �14

�

number of one kind of possible outcome

total number of all possible outcomes

Probability

Figure 10 Probability ofheads or tails. The probabilitythat a tossed coin will land on heads is �

12

�. The probabilitythat a tossed coin will land ontails is �

12

�.

173

Using the FigureMake sure students understandthat the probabilities in each squarein Figure 11 were obtained by mul-tiplying the probability at the topof the box by the probability alongthe side of the box. TAKS 1 Bio/IPC 2C

GENERAL



MISCONCEPTION ALERT

Probabilities Students may think thatprobabilities in genetic crosses show thedefinite outcome of a genetic cross. Pointout that probabilities are used only to predictthe possible outcome of a genetic cross.

Predicting theResults of CrossesUsing ProbabilitiesSkills AcquiredCalculating, applyinginformation

Teacher’s NotesHave students set up Punnettsquares similar to those inFigures 8 and 9. Then ask themto write in the probabilities offinding a specific allele in agamete.

Answers to Analysis 1. �

14

� 2. �12

�

3. 1 4. 0<x + 6x - 7 - 0

2

18

49376

0

52

Writing Skills Have studentsdevelop stories from which a pedi-gree can be drawn. Encourage themto be creative in thinking of charac-ters and traits that they choose tofollow through several generations.To illustrate their pedigree stories,student can add “family portraits.”Read some of the stories in class,and have students draw the pedi-grees from the information given in each story. VerbalLS

BUILDERSKILL

CulturalAwarenessCulturalAwareness

Albinism in Hopi Tribes A survey of aHopi tribe in Arizona found the frequencyof albinism to be 1 in 277. In contrast,albinism is very rare or nonexistent in otherNative American communities in Arizonaand New Mexico. Why is the frequency sohigh among the Hopi? The Hopi peoplehave always had a high regard for albinosand clan leaders have taken special care toprotect them from the harsh desert sun. Thistype of selection could explain the increasein albinism in the community.TAKS 2 Bio 6D, TAKS 3 Bio 7B

HeadsHeads

1_4

TailsTails

1_4

Heads1_2

Heads 1_2

Tails 1_2

Tails1_2

HeadsTails

1_4

TailsHeads

1_4

The green boxes have the samecombination (heads, tails), sothe probabilities are addedtogether.

1_4

+ 1_4

= 1_2

Probability of eachcoin landing onheads or tails

The possible results of tossing a nickel and apenny at the same time and the probability ofeach outcome are shown in Figure 11. Since thecombination of heads and tails can occur in twopossible ways, those two probabilities are addedtogether.

�14

� � �14

� � �24

� or �12

�

Consider the possible results that can occurin a cross between two pea plants that are het-erozygous for seed shape (Rr). The R allele forround seed shape is dominant over the r allelefor wrinkled seed shape. The probability of eachparent carrying gametes with R or r alleles is �

12

�.The probability of offspring with RR alleles is

�12

� � �12

� � �14

�

Similarly, the probability of offspring with rralleles is

�12

� � �12

� � �14

�

The combination of Rr alleles can occur in twopossible ways. One parent can contribute the Rallele, and the second parent the r allele, or viceversa. Thus, the probability of offspring with Rralleles is

�14

� � �14

� � �12

�

Analysis

1. Calculate the probability ofhomozygous dominant (BB )offspring resulting from across between two heterozy-gous (Bb) parents.

2. Calculate the probability ofheterozygous offspring result-ing from a cross between a

heterozygous parent and ahomozygous recessive (bb)parent.

3. Calculate the probability ofheterozygous offspring result-ing from a cross between ahomozygous dominant parentand a homozygous recessive parent.

4. Calculate the probability ofhomozygous dominant off-spring resulting from a crossbetween a heterozygous parent and a homozygousrecessive parent.

Predicting the Results ofCrosses Using ProbabilitiesBackground

In rabbits, the allele B for black hair is dominantover the allele b for brown hair. You can practiceusing probabilities to predict the outcome of geneticcrosses by completing the genetic problems below.Draw Punnett squares for each problem.

<x + 6x - 7 - 02

8

493 0

52

The probability of the results of flipping two coinsis easy to compute.

Figure 11 Probability with two coins

2C 6D

174

TAKS 1 Bio/IPC 2C, TAKS 2 Bio 6D

Student Edition TAKS Obj 1 Bio/IPC 2C TAKS Obj 2 Bio 6A TAKS Obj 2 Bio 6C TAKS Obj 2 Bio 6D TEKS Bio 6A, 6C, 6DTEKS Bio/IPC 2C

Teacher Edition TAKS Obj 1 Bio/IPC 2C TAKS Obj 2 Bio 6A, 6D TAKS Obj 3 Bio 7B TEKS Bio 3F, 6A, 6D, 7BTEKS Bio/IPC 2C

pp. 174–175

TAKS 1, TAKS 2

Math TAKS Obj 9, 8.11A; Obj 10, 8.14A

Vocabulary Ask students to differ-entiate between Punnett squares,probabilities, and pedigrees.(Punnett squares predict the expectedoutcome of a cross by considering allpossible combinations of gametes in a cross. Probabilities predict themathematical likelihood that a spe-cific event, such as the outcome of across, will occur. Pedigrees provide avisual representation of how a trait isinherited over several generations.)

Teaching TipSex linked Tell the students thatsome traits are not inherited equallyby both sexes. A sex-linked trait isusually seen only in males, and mostare recessive. Ask students how amale might inherit a sex-linkedtrait from his mother. (The mothercarries the trait as a recessive on oneof her X-chromosomes; the soninherits this chromosome from hismother and a Y from his father.)

Using the Figure Before teaching students to interpreta pedigree such as that shown inFigure 12, introduce the symbols:male (square), female (circle), traitexpressed (shaded circle or square),and trait not expressed (circle orsquare not shaded). Once studentsare comfortable with the meaningsof the symbols, have them interpretthe pedigree in Figure 12. Tell stu-dents the gene for this trait not onlyresults in a deficiency of skin, hair,and eye pigmentation but also causesdefects in vision. VisualTAKS 1 Bio/IPC 2C

LS

GENERAL

TAKS 2 Bio 6A

GENERALBUILDERSKILL


HISTORYHISTORYCONNECTIONCONNECTION

There is a high frequency of hemophiliaamong members of the royal familiesthroughout Europe. Queen Victoria was acarrier of sex-linked hemophilia. Becausemembers of the European nobility usuallymarried within their own social class, thehemophilia gene was passed via QueenVictoria’s daughters to the Russian, German,and Spanish royal families, increasing the fre-quency of the recessive allele amongEuropean nobility. Bio 3F, TAKS 2 Bio 6A

AnswerSome pedigreed dogs that areprone to genetic diseases includeIrish setters (blindness), Germanshepherds (hip dysplasia) anddaschounds (dwarfism).

TAKS 2 Bio 6C

Real LifeReal Life

In the wild, albino animals have little chance of survival. They lackthe pigments that provide protec-tion from the sun’s ultraviolet rays.

Inheritance of Traits Imagine that you want to learn about an inherited trait present inyour family. How would you find out the chances of passing the traitto your children? Geneticists often prepare a , a familyhistory that shows how a trait is inherited over several generations.Pedigrees are particularly helpful if the trait is a genetic disorderand the family members want to know if they are carriers or if theirchildren might get the disorder. Carriers are individuals who areheterozygous for an inherited disorder but do not show symptomsof the disorder. Carriers can pass the allele for the disorder to their offspring.

Figure 12 shows an example of a pedigree for a family withalbinism. In the genetic disorder albinism, the body is unable to pro-duce an enzyme necessary for the production of melanin. Melaninis a pigment that gives dark color to hair, skin, scales, eyes, andfeathers. Without melanin, an organism’s surface coloration may bemilky white and its eyes may be pink, as shown in Figure 12.

Scientists can determine several pieces of genetic informationfrom a pedigree:

Autosomal or Sex-Linked? If a trait is autosomal, it will appear inboth sexes equally. Recall that an autosome is a chromosome otherthan an X or Y sex chromosome. If a trait is sex-linked, it is usuallyseen only in males. A is a trait whose allele is locatedon the X chromosome. Most sex-linked traits are recessive. Becausemales have only one X chromosome, a male who carries a recessiveallele on the X or Y chromosome will exhibit the sex-linked condition.

A female who carries a recessive allele on one X chromosome willnot exhibit the condition if there is a dominant allele on her other Xchromosome. She will express the recessive condition only if sheinherits two recessive alleles. Thus, her chances of inheriting andexhibiting a sex-linked condition are significantly less.

sex-linked trait

pedigree

Female Female albino

Male Male albino

Horizontal linesindicate matings.

Vertical lines indicate offspring (arranged fromleft to right in order of their birth).

The purplesymbols representaffected individuals.

Albinism is a genetic disorder transmitted by a recessive allele.

Figure 12 Albinism pedigree

About 10 percent of Dalmatians are deaf.Because many purebred dogs are inbred—that is, they have closely related parents—some ofthem are homozygous forcertain recessive disorders.Finding Information If you have a purebred dog, find out if that breed is prone to a genetic disorder. 6C

175

TAKS 2

ReteachingWrite the following genotypes onthe board: (1) PP, (2) Pp, and (3)pp. Pair each student with a part-ner. Have students choose two ofthe genotypes and construct andcomplete a Punnett square showingthe cross. Have them share theirresults with their partners.

Quiz1. What is the probability of

two parents each carrying arecessive gene for an inheriteddisease to produce a child thatwill have that disease? (�

14

�)

2.Explain how the parents of anindividual with a recessive traitcan both be dominant for thattrait. (Parents are both heterozy-gous dominant.)

GENERAL

CloseClose

Answers to Section Review1. 3 freckles:1 no freckle; 1FF:2Ff:1ff

2. If, after a testcross, all of the offspring haveround seeds, the parent of the unknown islikely to be homozygous dominant. If, after atest cross, any of the offspring have wrinkledseeds, the parent with the unknown genotype is likely to be heterozygous.

3. �12

�

4. An individual will be a carrier if one parent of the individual is homozygous recessive, theother parent does not express the trait, and

TAKS 1 Bio/IPC 2C, TAKS 2 Bio 6ATAKS 2 Bio 6D (grade 10 only)

TAKS 1 Bio/IPC 2C, TAKS 2 Bio 6A, 6D (grade 10 only)

the individual in question does not express the trait.

5. A. Correct. This cross wouldproduce about �

14

� green seeds. B. Incorrect. This cross would not produce any green seeds.C. Incorrect. This cross would not produce anyyellow seeds. D. Incorrect. This cross wouldnot produce any green seeds.TAKS 1 Bio/IPC 2C, TAKS 2 Bio 6D (grade 10 only)

TAKS 1 Bio/IPC 2C


Evaluating aPedigreeSkills AcquiredAnalyzing, interpreting, drawing conclusions,applying information

Teacher’s NotesEncourage students to use “If-then” statements to organ-ize their thoughts and interpretthe pedigree. Example: If a traitis expressed by an offspring butnot by either parent, then thetrait must be recessive.

Answers to Analysis 1. autosomal recessive2. homozygous3. �

12

�

TAKS 1 Bio/IPC 2CTAKS 2 Bio 6D (grade10 only)

010001011001110101000100100010011100100100010000010100100111010101001000101010010010

Analysis

1. Interpret the pedigree todetermine whether the trait issex-linked or autosomal andwhether the trait is inherited in adominant or recessive manner.

2. Determine whether Female Ais homozygous or heterozygous.

3. Critical Thinking ApplyingInformation If Female B haschildren with a homozygousindividual, what is the probabil-ity that the children will beheterozygous?

Evaluating a Pedigree Background

The photo shows a family with an albino member.Pedigrees, such as the one below, can be used to trackdifferent genetic traits, including albinism. Use the pedi-gree below to practice interpreting a pedigree.

010001011001110101000100100010011100100100010000010100100111010101001000101010010010

Predict the expected phenotypic and genotypicratios among the offspring of two individuals whoare heterozygous for freckles (Ff ) by using aPunnett square. 2C 6A 6D

Summarize how a test cross can reveal the genotype of a pea plant with round seeds.

Calculate the probability that an individualheterozygous for a cleft chin (Cc) and an individ-ual homozygous for a cleft chin (cc) will produceoffspring that are homozygous recessive for acleft chin. (cc) 2C 6A

Critical Thinking Analyzing Graphics Whenanalyzing a pedigree, how can you determine if anindividual is a carrier (heterozygous) for the traitbeing studied? 2C

A cross between two peaplants that produce yellow seeds results in 124 off-spring: 93 produce yellow seeds and 31 producegreen seeds. What are the likely genotypes of theplants that were crossed? 2C 6D

A both Yy C both yyB both YY D one YY, one Yy


Section 3 Review

Dominant or Recessive? If the trait is autosomal dominant, everyindividual with the trait will have a parent with the trait. If the trait isrecessive, an individual with the trait can have one, two, or neitherparent exhibit the trait.

Heterozygous or Homozygous? If individuals with autosomal traitsare homozygous dominant or heterozygous, their phenotype willshow the dominant characteristic. If individuals are homozygousrecessive, their phenotype will show the recessive characteristic.Two people who are heterozygous carriers of a recessive mutationwill not show the mutation, but they can produce children who arehomozygous for the recessive allele.

Male Female

Female A

Male withtrait

Female withtrait

Female B

Albino

2C 6D

6D

176

Student Edition TAKS Obj 1 Bio/IPC 2C TAKS Obj 2 Bio 6A TAKS Obj 2 Bio 6C TAKS Obj 2 Bio 6D TEKS Bio 6A, 6C, 6DTEKS Bio/IPC 2C

Teacher Edition TAKS Obj 1 Bio/IPC 2C, 3A TAKS Obj 2 Bio 6A, 6D TEKS Bio 6A, 6DTEKS Bio/IPC 2C, 3A

pp. 176–177

TAKS 1, TAKS 2

Section 4

OverviewBefore beginning this sectionreview with your students theobjectives listed in the StudentEdition. Some traits are controlledby several genes, alleles may beequally dominant, or may be influ-enced by the environment or mutations. This section discussesthe inheritance of complex patterns of inheritance such as incompletedominance, codominance, polygenictraits, mutations and environmentalinfluences.

Ask students to study the animalsshown in Figure 16. Ask them tolist possible mechanisms that allowthe arctic fox to change its furcolor with changing seasons. (The temperature triggers enzymesinvolved in hormonal responses thatinfluence the genes.)

Discussion/Question Ask students to look at the variations in human traits as showin Figure 13. Ask them to proposea mechanism for the inheritance ofa trait such as eye color in humans,which can appear as brown, green,blue and gray. (There are at leastthree genes involved, brown, greenand blue; with brown dominant togreen and blue, and green dominantto blue.) Visual TAKS 2 Bio 6ALS

MotivateMotivate

Bellringer

FocusFocus


GENERAL

• Lesson Plan• Directed Reading• Active Reading GENERAL

GENERAL

Chapter Resource File

• Reading Organizers• Reading Strategies• Supplemental Reading Guide

The Double Helix

Planner CD-ROM

Transparencies

TT BellringerTT Some Human Genetic Disorders

A-Head 2-line

Section #

Objectives● Identify five factors that

influence patterns of heredity.

● Describe how mutations can cause genetic disorders.

● List two genetic disorders,and describe their causesand symptoms.

● Evaluate the benefits of genetic counseling.

Key Terms

polygenic traitincomplete dominancemultiple allelescodominance

Complex Patterns of Heredity

Complex Control of Traits A horse with red hair mates with a horse with white hair, and theiroffspring has both red and white hair. How can this be? If traits arecontrolled by single genes with simple dominant and recessive alleles, the colt’s hair should be one color or the other. Not always!Most of the time, traits, such as hair color in horses, display more-complex patterns of heredity than the simple dominant-recessivepatterns discussed so far.

Traits Influenced by Several Genes When several genes influence a trait, the trait is said to be a

. The genes for a polygenic trait may be scatteredalong the same chromosome or located on different chromosomes.Determining the effect of any one of these genes is difficult. Due toindependent assortment and crossing-over during meiosis, manydifferent combinations appear in offspring. Familiar examples ofpolygenic traits in humans include eye color, height, weight, andhair and skin color. All of these characteristics have degrees ofintermediate conditions between one extreme and the other, asshown in Figure 13.

Intermediate TraitsRecall that in Mendel’s pea-plant crosses, oneallele was completely dominant over another.In some organisms, however, an individualdisplays a trait that is intermediate betweenthe two parents, a condition known as

. For example, when asnapdragon with red flowers is crossed with asnapdragon with white flowers, a snapdragonwith pink flowers is produced. Neither thered nor the white allele is completely domi-nant over the other allele. The flowers appearpink because they have less red pigment thanthe red flowers. In Caucasians, the child of astraight-haired parent and a curly-haired par-ent will have wavy hair. Straight and curlyhair are homozygous dominant traits. Wavyhair is heterozygous and is intermediatebetween straight and curly hair.

incomplete dominance

polygenic trait

Section 4

Figure 13 Polygenic traits. Many traits—height, weight, hair color, and skin color—aretraits that are influenced by many genes.

6D

6C

6C

6C

177

IPC Benchmark Fact

Evaluate students’ ability to analyze, review, and cri-tique scientific explanations by asking them to identifyand describe the limitations of Mendel’s understandingof inheritance based on his pea plant experiments.Complete this exercise by comparing and contrastingsimple patterns of trait inheritance associated withpea plants with more complex patterns of trait inheri-tance such as polygenic traits, incomplete dominance,codominance, and multiple alleles.TAKS 1 IPC 3A

TAKS 2

TAKS 2

TAKS 2

TAKS 2

Teaching TipIncomplete Dominance Ask students whether a plant breedercould produce only pink floweringsnapdragons by crossing pink-flowering snapdragons and white-flowering snapdragons. Lead students to understand thatsince all pink-flowering snapdrag-ons are heterozygous, mating apink-flowering snapdragon with a white-flowering one would producepink-flowering and white-floweringoffspring in a ratio of 1:1.

Demonstration To convey the concept of universaldonor and universal acceptor, set upfour flasks to represent each bloodtype, and label them appropriately.The flasks should contain the fol-lowing: “A” blood (water with redfood color); “B” blood (water withblue food color); “AB” blood(water with red and blue foodcolor); and “O” blood (water only).Take an empty beaker and pour“O” blood into it. Show studentsthat pouring “A” blood into thebeaker containing “O” blood will“contaminate” the “O” blood(change its color). Demonstrate thepossible mixtures and have stu-dents derive which blood types arecompatible with each other. Pointout that A blood and B blood eachcontain unique carbohydrates thatO does not, which is why O is auniversal donor, and AB is a univer-sal acceptor. VisualTAKS 2 Bio 6D (grade 10 only)

LS

GENERAL

TAKS 2 Bio 6A, 6D (grade 10 only)

TeachTeach


did you know?Human Inheritance Mendel’s work with garden-pea plants showed that the traits hestudied are controlled by single genes. Inhumans, single-factor inheritance has beenfound in about 600 recessively inherited traits,and in such dominant conditions as Hunting-ton’s disease. However, many more conditionsare determined by polygenic inheritance, whichinvolves several genes. Such conditions includecleft lip and palate, schizophrenia, hyperten-sion, and diabetes. Bio 3F, Bio/IPC 3C

Traits Controlled by Genes with Three or More AllelesGenes with three or more alleles are said to have

. For example, in the human population, the ABO bloodgroups (blood types) are determined by three alleles, IA, IB, andi. The letters A and B refer to two carbohydrates on the surfaceof red blood cells. In the i allele, neither carbohydrate is present.The IA and IB alleles are both dominant over i. But neither IA norIB is dominant over the other. When IA and IB are both presentthey are codominant. Even for traits controlled by genes withmultiple alleles, an individual can have only two of the possiblealleles for that gene. Figure 14 shows how combinations of thethree different alleles can produce four different blood types—A,B, AB, and O. Notice that a person who inherits two i alleles hastype O blood.

Traits with Two Forms Displayed at the Same TimeFor some traits, two dominant alleles are expressed at the sametime. In this case, both forms of the trait are displayed, a phe-nomenon called . Codominance is different fromincomplete dominance because both traits are displayed.

The situation of human ABO blood groups, as discussedabove, is an example of co-dominance. The genotype of a personwho has blood type AB is IAIB, and neither allele is dominantover the other. Type AB blood cells carry both A- and B-types ofcarbohydrate molecules on their surfaces.

codominance

allelesmultiple

Different combinations of the three alleles IA, IB, and i result in four different bloodphenotypes, A, AB, B, and O. For example, a person with the alleles IA and iwould have blood type A.

Figure 14 Multiple alleles control the ABO blood groups

IAIA IAIB IAi

IAIB IBIB IBi

IAi IBi ii

IA IB i

IA

IB

i

Possible alleles

Po

ssib

le a

llele

s

ABlood types AB B O

178

Student Edition TAKS Obj 1 Bio/IPC 2C TAKS Obj 2 Bio 6A TAKS Obj 2 Bio 6D TEKS Bio 6A, 6DTEKS Bio/IPC 2C

Teacher Edition TAKS Obj 1 Bio/IPC 2A, 2C, 2D, 3CTAKS Obj 1 IPC 9BTAKS Obj 2 Bio 4B, 6A, 6D TEKS Bio 3F, 4B, 6A, 6DTEKS Bio/IPC 2A, 2C, 2D, 3CTEKS IPC 9B

pp. 178–179

Discussion Ask students to lookat the girl having a blood test donein Figure 14. Lead a brief discus-sion on blood tests. Tell studentsthat testing for specific materials inthe blood can discover many disor-ders. For example: anemia (too fewred blood cells, thus test for redblood cell count), diabetes (inabil-ity to break down blood sugar, thustest for blood sugar levels), andhigh cholesterol (thus test for HDL).

Using the Figure Have the students study Figure 14.Explain how the table shows thepossible blood types, and the use of“I” with the subscripts A, B, AB todenote alleles dominant to i.

Activity Using Punnet Squares Ask stu-dents to use a Punnett square tofigure out the following problem: Ifthe Mother of a type O child is A,list the mother’s genotype and thepossible genotypes for the father.(Mother is IAi; possible genotypes forthe father are IAi, IBi, ii)

DemonstrationUse litmus paper to test a weak acidsuch as vinegar, and a weak basesuch as a baking soda solution. Havestudents note the color change, redto blue for base and blue to pink(red) for acid. Then have studentslook at the flowers in Figure 15. Askthem is they can relate the color ofthe flowers to the litmus test usedfor acids and bases. In the case ofthe Hydrangea, the flowers are blueif the soil is acidic. (Litmus is a dyemade from organisms called lichens.)Ask student how the Hydrangeacould be used as a bio-indicator forthe acidity of the soil. VisualTAKS 1 Bio/IPC 2C, 2D

LS

GENERAL


GENERAL

TAKS 2 Bio 4B, 6D (grade 10 only)

SKILLBUILDER

READINGREADING


GENERAL

The blood groups A, B, AB, and O all have an identical sugar chain on their cell surface.Type A cells have an additional sugar, type Bcells have a different additional sugar, andtype AB cells have both additional sugars.Researchers hope to produce type O cells byusing enzymes to remove the additional sugarsfrom type A, B, and AB cells. Ask students whythis would be an important medical break-through. (Any blood type could be converted totype O, which would make it compatible with allother blood types as a universal donor.) Bio/IPC 3C

MEDICINEMEDICINECONNECTIONCONNECTION

Traits Influenced by the EnvironmentAn individual’s phenotype often depends on condi-tions in the environment. In plants, hydrangea (hieDRAYN juh) flowers of the same genetic varietyrange in color from blue to pink, as shown in Figure 15. Hydrangea plants in acidic soil bloomblue flowers, while those in neutral to basic soil willbloom pink flowers.

The color of the arctic fox is affected by tempera-ture. During summer, the fox produces enzymes thatmake pigments. These pigments darken the fox’s coatto a reddish brown, as shown in Figure 16, enablingthe fox to blend in with the summer landscape.During the winter, the pigment-producing genes ofthe arctic fox do not function because of the cold tem-perature. As a result, the coat of the fox is white, andthe animal blends in with the snowy background.

Fur color in Siamese cats is also influenced by temperature. In aSiamese cat, the fur on its ears, nose, paws, and tail is darker thanon the rest of its body. The Siamese cat has a genotype that resultsin dark fur at locations on its body that are cooler than the normalbody temperature. Thus, the darkened parts have a lower body tem-perature than the light parts.

In humans many traits, such as height, are influenced by the envi-ronment. For example, height is influenced by nutrition, an internalenvironmental condition. Exposure to the sun, an external environ-mental condition, alters the color of the skin. Many aspects of humanpersonality, such as aggressive behavior, are strongly influenced bythe environment, although genes appear to play an important role.Because identical twins have identical genes, they are often used tostudy environmental influences. Because identical twins are geneti-cally identical, any differences between them are attributed toenvironmental influences.

Can the same species of fox look so different? Many arctic mammals, such asthe arctic fox, develop white fur during the winter and dark fur during the summer.

Figure 16 Environmental influences on fur color

Figure 15 Environmentalinfluences on flower color.Hydrangea with the samegenotype for flower colorexpress different phenotypesdepending on the acidity of the soil.

179

IPC Benchmark Fact

Review the pH scale with students and have themdesign an experiment to test the effects of various pHenvironments on hydrangea. Be sure to have themidentify the experimental hypothesis, which accordingto the textbook is that an acidic environment producesblue flowers while a neutral or basic environmentproduces pink flowers. If time permits, have studentsconduct the experiment in the laboratory in order todemonstrate how chemistry—pH in this instance—affects the everyday physical expression of a trait of phenotype. TAKS 1 Bio/IPC 2A; TAKS 4 IPC 9B

Discussion/Question Ask students why most bald peopleare male. Tell them both males andfemales can inherit a “baldness”allele. (The male hormone testosteroneactivates the allele and eventually leadsto baldness. Women produce smallamounts of testosterone. However,baldness does not occur in femalesunless they have both alleles for baldness. The presence of only oneallele for baldness causes men tobecome bald.)

Teaching TipOne faulty gene can alter ahemoglobin molecule.Hemoglobin is a protein. A mutationin a hemoglobin gene that results in achange in the amino acid sequence ofthe gene can alter the structure of theprotein and ultimately the protein’sfunction.

Teaching TipSickle Cell Anemia and MalariaExplain to students the adaptivevalue in individuals that have onegene for sickle cell anemia. Askthem how this might explain thehigher incidence of the disease inAfrican Americans. (The adaptivevalue is a less severe affect frommalaria. Since malaria is more prevalent in Africa, those individualsinheriting one allele for sickle cellanemia are somewhat protectedagainst malaria, thus retaining thegene in the population, as comparedto those with two alleles for sicklecell anemia, who would succumb tothe disease, lessening the chance ofpassing on the gene.) TAKS 3 Bio 7B

TAKS 2 Bio 4B

TAKS 2 Bio 6A



GENERAL

HISTORYHISTORYCONNECTIONCONNECTION

Working with limited laboratory facilities anda strong determination to fight the disease thatwas killing their friends and families, twoAfrican-American researchers, Dr. AngelaFerguson and Dr. Roland Scott, published apaper on sickle cell anemia in the 1940’s—25 years ahead of other researchers. Dr. Scott, known as the “father of sickle cellanemia research,” is the founder and formerdirector of Howard University’s Center forSickle Cell Anemia Research. Dr. Fergusonwas an associate professor of pediatrics atHoward University. Bio 3F

Genetic Disorders In order for a person to develop and function normally, the proteinsencoded by his or her genes must function precisely. Unfortunately,sometimes genes are damaged or are copied incorrectly, resultingin faulty proteins. Changes in genetic material are called mutations.Mutations are rare because cells have efficient systems for correct-ing errors. But mutations sometimes occur, and they may haveharmful effects.

The harmful effects produced by inherited mutations are calledgenetic disorders. Many mutations are carried by recessive allelesin heterozygous individuals. This means that two phenotypicallynormal people who are heterozygous carriers of a recessive muta-tion can produce children who are homozygous for the recessiveallele. In such cases, the effects of the mutated allele cannot be avoided. Several human genetic disorders are summarized inTable 2.

Sickle Cell Anemia An example of a recessive genetic disorder is sickle cell anemia, acondition caused by a mutated allele that produces a defectiveform of the protein hemoglobin. Hemoglobin is found within redblood cells, where it binds oxygen and transports it through thebody. In sickle cell anemia, the defective form of hemoglobincauses many red blood cells to bend into a sickle shape, as seen inFigure 17. The sickle-shaped cells rupture easily, resulting in lessoxygen being carried by the blood. Sickle-shaped cells also tendto get stuck in blood vessels; this can cut off blood supply to an organ.

The recessive allele that causes sickle-shaped red bloodcells also helps protect the cells of heterozygous individualsfrom the effects of malaria. Malaria is a disease caused bya parasitic protozoan that invades red blood cells. The sick-led red blood cells of heterozygous individuals cause thedeath of the parasite. But the individual’s normal red bloodcells can still transport enough oxygen. Therefore, thesepeople are protected from the effects of malaria thatthreaten individuals who are homozygous dominant for thehemoglobin gene.

Cystic Fibrosis (CF) Cystic fibrosis, a fatal recessive trait, is the most commonfatal hereditary disorder among Caucasians. One in 25 Cau-casian individuals has at least one copy of a defective genethat makes a protein necessary to pump chloride into and outof cells. About 1 in 2,500 Caucasian infants in the UnitedStates is homozygous for the cf allele. The airways of thelungs become clogged with thick mucus, and the ducts of theliver and pancreas become blocked. While treatments canrelieve some of the symptoms, there is no known cure.

Figure 17 Sickle cell. Oneout of 500 African Americanshas sickle cell anemia, which is caused by a gene mutationthat produces a defective formof hemoglobin.

Magnification: 13,6003�

180

Student Edition TAKS Obj 2 Bio 6A TAKS Obj 2 Bio 6C TAKS Obj 2 Bio 6D TEKS Bio 6A, 6C, 6D

Teacher Edition TAKS Obj 2 Bio 4B, 6ATAKS Obj 3 Bio 7B TAKS Obj 1 Bio/IPC 2C TEKS Bio 3F, 4B, 6A, 7BTEKS Bio/IPC 2C

pp. 180–181

Group Activity Patterns of Heredity Pair stu-dents and ask each pair to make atable to organize information aboutpatterns of heredity that are morecomplex than simple dominant-recessive patterns. The studentsshould write the following headingsacross the top: Explanation,Example(s). Along the sides, stu-dents should write the following:Polygenic Traits, Incomplete domi-nance, Codominance, Multiple alleles, and Environmentally influ-enced traits. Have students addinformation to the table as theyreview this section. Logical

Reading Organizer Have studentsmake a reading organizer describingthe cause and effect of each of thegenetic disorders discussed in thissection. Students should construct acause effect graph for each disease.

Logical TAKS 1 Bio/IPC 2CLS

SKILLBUILDER

READINGREADING

TAKS 1 Bio/IPC 2CLS

GENERAL


Women who have PKU often have babies withmental retardation, not because the baby hasPKU, but because the mother’s body chemistryis altered during pregnancy. These babies can-not be helped with special diet. However, themental retardation can be avoided if themother follows a low-phenylalanine dietbefore and during pregnancy.

REAL WORLDREAL WORLDCONNECTIONCONNECTION

Witchcraft or Disease? In the United States,many cases of Huntington’s disease can betraced back to two brothers. The two menimmigrated to North America from England inthe 1600’s because of accusations of witchcraftin their family. The family members of these twobrothers were apparently persecuted because oftheir strange behaviors, which are now under-stood to be symptoms of Huntington’s disease.Constant dance-like movements in its victimscharacterize the disease. Bio 3F

Hemophilia Another recessive genetic disorder is hemophilia (hee moh FIHLee uh), a condition that impairs the blood’s ability to clot.Hemophilia is a sex-linked trait. More than a dozen genes code forthe proteins involved in blood clotting. A mutation on one of thesegenes on the X chromosome causes the form of hemophilia calledhemophilia A. If the mutation appears on the X chromosome,which a male receives from his mother, he does not have a normalgene on the Y chromosome to compensate. Therefore, he willdevelop hemophilia.

Huntington’s Disease (HD) Huntington’s disease is a genetic disorder caused by a dominantallele located on an autosome. The first symptoms of HD—mildforgetfulness and irritability—appear in victims in their thirties orforties. In time, HD causes loss of muscle control, uncontrollablephysical spasms, severe mental illness, and eventually death.Unfortunately, most people who have the HD allele do not knowthey have the disease until after they have had children. Thus, thedisease is unknowingly passed on from one generation to the next.

Table 2 Some Human Genetic Disorders

Disorder Dominant or Symptom Defect Frequency Among Recessive Human Births

Sickle Cell Recessive Poor blood circulation Abnormal hemoglobin 1:500 Anemia molecules (African Americans)

Hypercholes- Dominant Excessive cholesterol Abnormal form of cell 1:500terolemia levels in blood, leading surface receptor for

to heart disease cholesterol

Tay-Sachs Recessive Deterioration of central Defective form of a 1:3,500 Disease in early childhood nervous system; death brain enzyme (Ashkenazi Jews)

in early childhood

Cystic Recessive Mucus clogs organs Defective chloride-ion 1:2,500Fibrosis including the lungs, transport protein (Caucasians)

liver, and pancreas; affected individuals usually do not survive to adulthood

Hemophilia A Sex-linked Failure of blood Defective form of a 1:10,000(Classical) recessive to clot blood-clotting factor (males)

Huntington’s Dominant Gradual deterioration Inhibitor of brain-cell 1:10,000Disease of brain tissue in middle metabolism is made

age; shortened lifeexpectancy

www.scilinks.orgTopic: Genetic DisordersKeyword: HX4091

181

ReteachingK-W-L Tell students to return totheir list of things they want to knowabout inheritance, which they cre-ated in Section 1. Have them placecheck marks next to the questionsthat they are now able to answer.Students should finish by making a list of what the have Learned.Conclude by asking students whichquestions are still unanswered. Askif they have new questions.

QuizGive an example of a possible trait for each of the following genotypic conditions:

1. Incomplete dominance (flowercolor in snapdragons or hair shapein humans)

2.Multiple alleles (blood type inhumans)

3. Codominance (roan coat inhorses) disorder (sickle cell anemia,Tay-Sachs or cystic fibrosis).

AlternativeAssessmentAssign pairs of students a geneticdisorder and have the studentsdesign an informative brochureabout the disorder, similar tobrochures found in a doctor’soffice. Set guidelines on informationyou expect students to include, suchas symptoms, causes, prognosis andsupport groups. Sample brochurescan be obtained from variousmedical facilities. Have studentspresent their brochures to the class. TAKS 1 Bio/IPC 2C

Co-op Learning

GENERAL

CloseClose

Answers to Section Review

1. Incomplete dominance produces traits that areintermediate between two contrasting forms ofa trait. In codominance, both dominant formsof a trait are displayed at the same time.

2. Answers will vary but may include fur color inSiamese cats and arctic mammals, flower colorin hydrangea plants, and height and skin colorin humans.

3. A genetic disorder results when a mutation isinherited and the mutation produces harmfuleffects. TAKS 2 Bio 6C, 6D (grade 10 only)


4. The male receives from his mother an X chro-mosome with a mutated blood-clotting gene.

5. Students should agree. It would require that anindividual have three alleles—IA, IB, and i.

6. A. Incorrect. Hemophilia A is an example of a sex-linked recessive allele. B. Incorrect. Huntington’s is dominant but not sex-linked. C. Incorrect. Sickle cell anemia,Tay-Sachs and cystic fibrosis are examples of autosomal recessive alleles. D. Correct.Huntington’s disease is an example of auto-somal dominant allele. TAKS 2 Bio 6C


TAKS 2 Bio 6C, 6D (grade 10 only)


Differentiate between incomplete dominanceand codominance.

Identify two examples of traits that areinfluenced by environmental conditions.

Summarize how a genetic disorder can resultfrom a mutation. 6C 6D

Describe how males inherit hemophilia.

Critical Thinking Justifying ConclusionsA nurse states that a person cannot have the bloodtype ABO. Do you agree or disagree? Explain.

The mutated allele that causesHuntington’s disease is A sex-linked and C autosomal and

recessive. recessive.B sex-linked and D autosomal and

dominant. dominant.


Section 4 Review

www.scilinks.orgTopic: Genetic CounselingKeyword: HX4090

Treating Genetic Disorders Most genetic disorders cannot be cured, although progress is beingmade. A person with a family history of genetic disorders may wishto undergo genetic counseling before becoming a parent. Geneticcounseling is a form of medical guidance that informs people aboutgenetic problems that could affect them or their offspring.

In some cases, a genetic disorder can be treated if it is diagnosedearly enough. For example, an individual with the genetic disorderphenylketonuria (PKU) lacks an enzyme that converts the aminoacid phenylalanine into the amino acid tyrosine. As a result, pheny-lalanine builds up in the body and causes severe mental retardation.If PKU is diagnosed soon after birth, however, the newborn canbe placed on a low-phenylalanine diet. Because this disorder can be easily diagnosed by inexpensive laboratory tests, many states requirePKU testing of all newborns.

Gene Therapy Gene technology may soon allow scientists to correct certain recessivegenetic disorders by replacing defective genes with copies of healthyones, an approach called gene therapy. The essential first step in genetherapy is to isolate a copy of the gene. The defective cf gene was iso-lated in 1989. In 1990, a working cf gene was successfully transferredinto human lung cells growing in tissue culture by attaching the cfgene to the DNA of a cold virus. The cold virus—carrying the normalcf gene piggyback—easily infects lung cells. The cf gene enters the lungcells and begins producing functional CF protein. Thus, the defectivecells are “cured” and are able to transport chloride ions across theirplasma membranes.

Similar attempts in humans, however, were not successful. Mostpeople have had colds and, as a consequence, have built up a naturalimmunity to the cold virus. Their lungs therefore reject the cold virusand its cf passenger. In the last few years, similar attempts using adifferent virus to transport the cf gene into lung cells have been ini-tiated. This virus, called AAV, produces almost no immune responseand so seems a much more suitable vehicle for introducing cf intocells. Clinical trials are underway, and the outlook is promising.

6C

6D6D

6C 6D

182

Student Edition TAKS Obj 2 Bio 6C TAKS Obj 2 Bio 6D TEKS Bio 6C, 6D

Teacher Edition TAKS Obj 1 Bio/IPC 2C TAKS Obj 2 Bio 6C, 6D TAKS Obj 4 IPC 9A, 9B TEKS Bio 6C, 6D TEKS Bio/IPC 2C

pp. 182–183

Alternative Assessment Assign each student one of thefollowing topics: incomplete domi-nance, co-dominance, or multiplealleles. Have each student think of aconcrete example for teaching theirassigned topic to others. Have eachstudent give a short oral report,using their chosen example toexplain their topic.

GENERAL

Answer to Concept Map

The following is one possible answer toPerformance Zone item 15.


• Science Skills Worksheet• Critical Thinking Worksheet• Test Prep Pretest• Chapter Test GENERAL

GENERAL

GENERAL

Chapter Resource File

self-pollinated

to produce two

cross-pollinated to produce

self-pollinated to produce

P generations

F1 generation

F2 generation

Pisum sativum

described

of

which can code for a

alleles

segregationindependentassortment

dominanttrait

recessivetrait

Mendel

Key Concepts

Study CHAPTER HIGHLIGHTS

ZONE

The Origins of Genetics● Gregor Mendel bred varieties of the garden pea in an

attempt to understand heredity. Mendel observed that con-trasting traits appear in offspring according to simple ratios.

● In Mendel’s experiments, only one of the two contrastingforms of a trait was expressed in the F1 generation. The otherform reappeared in the F2 generation in a 3:1 ratio.

Mendel’s Theory● Different versions of a gene are called alleles. An individual

usually has two alleles for a gene, each inherited from a different parent.

● Individuals with the same two alleles for a gene are homozygous; those with two different alleles for a gene are heterozygous.

● The law of segregation states that the two alleles for a trait separate when gametes are formed. The law of independentassortment states that two or more pairs of alleles separateindependently of one another during gamete formation.

Studying Heredity ● The results of genetic crosses can be predicted with the use

of Punnett squares and probabilities.● A test cross can be used to determine whether an individual

expressing a dominant trait is heterozygous or homozygous.● A trait’s pattern of inheritance within a family can be deter-

mined by analyzing a pedigree.

Complex Patterns of Heredity● Traits usually display complex patterns of heredity, such as

incomplete dominance, codominance, and multiple alleles.● Mutations can cause genetic disorders, such as sickle cell

anemia, hemophilia, and Huntington’s disease.● Genetic counseling can help patients concerned about a

genetic disorder.

4

3

2

1

Key Terms

Section 1heredity (162)genetics (162)monohybrid cross (164)true-breeding (164)P generation (164)F1 generation (164)F2 generation (164)

Section 4polygenic trait (177)incomplete dominance (177)multiple alleles (178)codominance (178)

Section 2allele (167)dominant (167)recessive (167)homozygous (167)heterozygous (167)genotype (168)phenotype (168)law of segregation (169)law of independent assortment

(169)

Section 3Punnett square (170)test cross (172)probability (173)pedigree (175)sex-linked trait (175)

BIOLOGYBIOLOGYUnit 5—Heredity Use this unit to review the key concepts and terms in this chapter.

183

IPC BenchmarkReview

To prepare students for the TAKS, havestudents review Solution Chemistry:Water as a Universal Solvent andConcentrations of Solutions TAKS Obj 4IPC 9A, 9B on pp. 1053–1054 of the IPC Refresher in the Texas AssessmentAppendix of this book.

ANSWERS

Using Key Terms1. a2. c3. d4. a5. a. A dominant trait appears in a

heterozygous individual; arecessive trait is hidden in a het-erozygous individual.

b. Homozygous refers to an indi-vidual with two identical allelesfor a trait. Heterozygous refersto an individual with two dif-ferent alleles for a trait.

c. The law of segregation statesthat the two alleles for a traitseparate when gametes areformed. The law of independentassortment states that the allelesof different genes separate independently of one anotherduring gamete formation.

Understanding Key Ideas

6. b7. b8. b9. a

10. c11. c12. A copy of the functional gene is

attached to the DNA of a virus.The functional gene gets into thedefective cells by “piggybacking”on the virus. Once inside thecells, it produces a functionalprotein that helps remedy the disease. TAKS 2 Bio 6A, 6D

(grade 10 only)

TAKS 2 Bio 6D (grade 10 only)TAKS 2 Bio 6C

TAKS 2 Bio 6D (grade 10 only)TAKS 2 Bio 6D (grade 10 only)Bio 3F

TAKS 2 Bio 6D (grade 10 only)TAKS 2 Bio 6D (grade 10 only)TAKS 2 Bio 6D (grade 10 only)TAKS 2 Bio 6D (grade 10 only)

13. During meiosis II, the members of each pairof alleles separate when gametes are formedas described in the law of segregation.

14. 1 YYRR : 2 YyRR : 1 yyRR; 3 yellow, round : 1 green, round.

15. One possible answer to the concept map isfound at the bottom of the Study Zone page.TAKS 2 Bio/IPC 2C, Bio 3E


Bio 6E


CHAPTER 8

Section Questions1 1, 2, 6, 7, 82 5, 9, 13, 15, 16, 203 3, 10, 11, 144 4, 12, 17, 18, 19, 21, 22

Assignment Guide

Review and Assess2C, 2D, 3C, 3D, 3E, 3F, 6A, 6C, 6D, 6E, 7B

Using Key Terms 1. The offspring of true-breeding parents are

called thea. F1 generation.b. F2 generation.c. dominant offspring.d. recessive offspring.

2. The color of a dog’s coat is the dog’sa. dominance. c. phenotype.b. pedigree. d. genotype.

3. The unknown genotype of an individualwith a dominant phenotype can be deter-mined usinga. a ratio. c. probability.b. a dihybrid cross. d. a test cross.

4. A trait with two dominant alleles that areexpressed at the same time isa. codominant.b. mutationalc. incompletely dominant.d. polygenic.

5. For each pair of terms, explain the differ-ences in their meanings.a. dominant, recessiveb. homozygous, heterozygousc. law of segregation, law of independent

assortment

Understanding Key Ideas6. The scientist whose studies formed the

basis of modern genetics isa. T. A. Knight. c. Louis Pasteur.b. Gregor Mendel. d. Robert Hooke.

7. Which of the following is not a good reasonwhy Pisum sativum makes an excellentsubject for genetic study?a. Many varieties exist.b. They require cross-pollination.c. They grow quickly.d. They demonstrate complete dominance.

8. If smooth peas are dominant over wrinkledpeas, the allele for smooth peas should berepresented asa. W. b. S. c. w. d. s.

9. The law of segregation states that pairs ofallelesa. separate when gametes form.b. separate independently of one another

during gamete formation.c. are always the same.d. are always different.

10. The trait shown below is

a. sex-linked and dominant.b. autosomal and dominant.c. sex-linked and recessive.d. autosomal and recessive.

11. D, dimples, is the dominant allele to therecessive allele, d, no dimples. The proba-bility of parents with Dd and dd genotypeshaving a child with no dimples (dd) is

a. �18

�. c. �12

�.

b. �14

�. d. 1.

12. Explain how working genes have beeninserted into defective cells during genetherapy.

13. Relate the events of meiosis to the law ofsegregation. (Hint: See Chapter 7, Section 1.)

14. State the genotypic andphenotypic ratios that would result from a cross between two YyRR pea plants.

15. Concept Mapping Make a conceptmap about Mendel’s experiments. Try toinclude the following words in your map:Pisum sativum, P generation, F1 generation,F2 generation, dominant trait, and recessivetrait.

PerformanceZONE

CHAPTER REVIEW

3F

6C

6A 6D

6E

2C 3E

6D

6D

6D

6D

6D

6D

6D

6D

184

Review and AssessTAKS Obj 1 Bio/IPC 2DTAKS Obj 2 Bio 6A, 6C, 6DTAKS Obj 3 Bio 7BTEKS Bio 3D, 3E, 3F, 6A, 6C, 6D, 6E,7BTEKS Bio/IPC 2C, 2D, 3C

pp. 184–185

Critical Thinking

16. Patterns obtained from largesamples are less likely to be dis-torted by rare events that canoccur by chance.

17. Since domesticated animals aremore likely to be inbred, manyare homozygous for many traitsand thus prone to inherited reces-sive traits such as albinism.

18. Cystic fibrosis is a recessive auto-somal disorder. Thus, each parentmust have the recessive allele.Changes are increased that hissister is a carrier (heterozygote)for cystic fibrosis.

19. Answers will vary, but studentsmight suggest that it mightcontribute to a resurgence ofbehavioral genetic determinism—the belief that genetics is the majorfactor in determining behavior.This might result in prejudiceagainst or for individuals with acertain genotype.

20. Reports and displays will vary.Gene technology is now used inplant breeding. Many plantbreeders use gene technologyequipment to conduct theirbreeding program.

21. Genetic counselors use varioustypes of information, includingpedigrees, laboratory tests, andkaryotypes, to determine theodds of a person or a couple’schild having a genetic disorder.Genetic counselors also outlinethe options for dealing with thoserisks and offer emotional support.Genetic counseling requires a spe-cialized graduate degree andexperience in the areas of medicalgenetics and counseling. Employersinclude university medical centers,private hospital settings, healthmaintenance organizations, andlaboratories. Growth prospectsare good. Starting salary will varyby region.

22. Answers will vary. Animal breed-ers use genetics to predict howoften a trait will appear whentwo animals are bred. Animalsbred for special purposes includedogs, cats, horses, goats, rabbits,and cattle.TAKS 1 Bio/IPC 2D, TAKS 2 Bio 6D(grade 10 only)

Bio 3D

TAKS 1 Bio/IPC 2D

Bio/IPC 3C

Bio/IPC 3C

TAKS 3 Bio 7B

TAKS 2 Bio 6C

1. A. Incorrect. This cross would produce all tt.B. Correct. Tt across the top and tt along theside would produce the arrangement shown in the Punnett square. C. Incorrect. This cross would produce 1 TT, 2 Tt, and 1 tt. D. Incorrect. This cross would produce all TT.

2. F. Correct. 2 Tt:2 tt, reduced to 1:1. G. Incorrect. Both parents would have to behybrid to produce this ratio. H. Incorrect. Toproduce any Tt, one parent has to have at least1 T, which, using the Punnett square, gives the


probability of 50% of the offspring inheritingT, not 25%. J. Incorrect. To produce any tt,each parent must have at least 1 t. If the par-ents are Tt, the ratio would be 1 TT:2 Tt:1 tt.If one parent was Tt and the other was tt, theratio would be 2 Tt:2 tt.

3. A. Incorrect. 0 would indicate that there are nott, when there are �

24

�. B. Incorrect. 60 would rep-resent only �

14

�; �12

� are tt. C. Correct. The Punnettsquare predicts �

12

� will be tt, or �12

� of 240, whichis 120. D. Incorrect. 180 represents �

34

�; �12

� are tt.TAKS 2 Bio 6D (grade 10 only)



Test

Critical Thinking16. Evaluating Results Mendel based his

conclusion about inheritance patterns onexperiments involving large numbers ofplants. Why do you think the use of largenumbers of individuals is advantageouswhen studying patterns of inheritance?

17. Inferring Relationships Albinism is rareamong wild animals but common amongsome domesticated species. What factorsmight account for this difference?

18. Justifying Conclusions A 20-year-old manwho has cystic fibrosis has a sister who isplanning to have a child. The man encour-ages his sister to see a genetic counselor.What do you think the man’s reasons arefor giving such advice?

19. Predicting Results How might researchthat demonstrates a genetic basis for someaspects of human behavior impact society?

Alternative Assessment20. Technology and Learning Find out how

new technologies have changed plant-breeding methods since Mendel’s time.Prepare an oral report to summarize your findings. Or create a display thatcompares the methods and equipmentMendel might have used with those usedby plant breeders today.

21. Career Connection Genetic CounselorResearch the field of genetic counseling,and write a report on your findings. Yourreport should include a job description,training required, kinds of employers,growth prospects, and starting salary.

22. Interactive Tutor Unit 5 Heredity Write areport summarizing how an understand-ing of heredity allows animal breeders todevelop animals with desirable traits.Find out what kinds of animals are bredfor special purposes.

The diagram below shows the expected resultsof a cross between two pea plants. T and trepresent the alleles for the tall and dwarf traits,respectively. Use the diagram and your knowl-edge of science to answer questions 1–3.

1. What are the genotypes of the plants thatwere crossed? A tt on the top; tt along the sideB Tt on the top; tt along the side C Tt on the top; Tt along the sideD TT on the top; TT along the side

2. What genotypic ratio is expected in theoffspring of this cross?F 1 Tt : 1 ttG 3 Tt : 1 ttH 1 Tt : 3 ttJ 1 TT : 1 tt

3. If this cross produced 240 offspring, howmany of the offspring would be expected tohave the dwarf trait?A 0B 60C 120 D 180

Scan the answer set for words such as “never” and“always.” Such words often are used in statementsthat are incorrect because they are too general.

Tt

tt

?

?

?

?

Tt

tt

The diagram below shows the expected resultsof a cross between two pea plants. T and trepresent the alleles for the tall and dwarf traits,respectively. Use the diagram and your knowl-edge of science to answer questions 1–3.

1. What are the genotypes of the plants thatwere crossed? A tt on the top; tt along the sideB Tt on the top; tt along the side C Tt on the top; Tt along the sideD TT on the top; TT along the side

2. What genotypic ratio is expected in theoffspring of this cross?F 1 Tt : 1 ttG 3 Tt : 1 ttH 1 Tt : 3 ttJ 1 TT : 1 tt

3. If this cross produced 240 offspring, howmany of the offspring would be expected tohave the dwarf trait?A 0B 60C 120 D 180.


7B

6C

3C

3C

6D

6D

6D

2D 3C

3D

2D 6D

185

Documents

Section 1 The Origins of Genetics - Jourdanton ISD · Before beginning this section review with your students the objectives listed in the Student ... TEKS Bio 3F, 6A, 6D, 10A Teacher