section 1 from genes to proteins - jourdanton · pdf file208 chapter 10 • how proteins...

Section 1

208 Chapter 10 • How Proteins Are Made

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

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GENERAL

GENERAL

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TT BellringerTT TranscriptionTT Codons in mRNATT Translation: Assembling Proteins

• Reading Organizers• Reading Strategies

Planner CD-ROM

OverviewBefore beginning this sectionreview with your students theobjectives listed in the StudentEdition. In this section, studentswill follow the events of transcrip-tion, during which information istransferred from a strand of DNAto a complementary strand ofRNA. This RNA is then used dur-ing translation, the process thatbuilds proteins. Students will learnhow the three-base codons ofmRNA are used to link specificamino acids in protein synthesis.

Ask students to name some bodyparts that contain proteins. (hair,skin, nails, internal organs) Point outthat chemical structures involved inphysiology are also made of pro-teins—hemoglobin in the blood,insulin that regulates blood glucoselevels, and enzymes that regulate allbody functions. Bio 9A

Bellringer

TAKS 2 Bio 6B (grade 11 only)

FocusFocus

• Unit 6—GeneExpression: Topics 3–6This engaging tutorial introduces stu-dents to principles of protein synthesiswithin the cell.

BIOLOGYBIOLOGY

StrategiesStrategiesINCLUSIONINCLUSION

To help students remember the steps of tran-scription and translation, have students writeeach step on an index card. The steps shouldbe numbered on the back of the index card.Students can mix the steps and re-order themfrom memory. When they have attempted toput the steps in order, they can turn the cardsover and check to see if they have them in thecorrect order. This can be a study guide foran individual student or small group.

• Learning Disability • English as a SecondLanguage

Section 1 From Genes to Proteins

Decoding the Information in DNATraits, such as eye color, are determined by proteins that are builtaccording to instructions coded in DNA. Recall that proteins havemany functions, including acting as enzymes and cell membranechannels. Proteins, however, are not built directly from DNA.Ribonucleic (rie boh noo KLAY ihk) acid is also involved.

Like DNA, is a nucleic acid—a moleculemade of nucleotides linked together. RNA differs from DNA in threeways. First, RNA consists of a single strand of nucleotides instead ofthe two strands found in DNA, as shown in Figure 1. Second, RNAnucleotides contain the five-carbon sugar ribose (RIE bohs) ratherthan the sugar deoxyribose, which is found in DNA nucleotides.Ribose contains one more oxygen atom than deoxyribose contains.And third, in addition to the A, G, and C nitrogen bases found inDNA, RNA nucleotides can have a nitrogen base called (YURuh sihl)—abbreviated as U. No thymine (T) bases are found in RNA.Like thymine, uracil is complementary to adenine whenever RNAbase-pairs with another nucleic acid.

A gene’s instructions for making a protein are coded in thesequence of nucleotides in the gene. The instructions for making aprotein are transferred from a gene to an RNA molecule in a processcalled . Cells then use two different types of RNA to readthe instructions on the RNA molecule and put together the aminoacids that make up the protein in a process called . Theentire process by which proteins are made based on the informationencoded in DNA is called , or protein synthesis. Thisprocess is summarized in Figure 1.

gene expression

translation

transcription

uracil

ribonucleic acid (RNA)

Objectives● Compare the structure

of RNA with that of DNA.

● Summarize the process oftranscription.

● Relate the role of codons tothe sequence of amino acidsthat results after translation.

● Outline the major steps oftranslation.

● Discuss the evolutionarysignificance of the geneticcode.

Key Terms

ribonucleic acid (RNA)uraciltranscriptiontranslationgene expressionRNA polymerasemessenger RNAcodongenetic codetransfer RNAanticodonribosomal RNA

The instructions for building a protein are found in a gene and are “rewritten” to a molecule of RNA during transcription. The RNA is then “deciphered” during translation.

Figure 1 Gene expression

DNA RNA ProteinTranscription Translation

UracilT

A

C

A

C

A

C

G

A

U

G

U

G

U

G

C

A

T

G

T

G

T

G

C

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Student Edition TAKS Obj 2 Bio 4B TAKS Obj 2 Bio 6A TAKS Obj 2 Bio 6B (grade 11only)TEKS Bio 4B, 6A, 6B

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pp. 208–209

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Discussion/QuestionDraw the structures of deoxyriboseand ribose on the board. Ask stu-dents to compare the two sugars.(Students should notice that thestructures are almost identical, exceptdeoxyribose has one less oxygen thanribose.) Ask students to explainhow this difference is reflected inthe names of the two sugars.(Deoxyribose is deoxygenated, relativeto ribose.)

VisualLS

MotivateMotivate

Chapter 10 • How Proteins Are Made 209

Some diseases, such as sickle cell anemia, are caused by changes in just one or a fewnucleotides of a gene. Contact a local labora-tory where genetic testing is performed, orcontact an organization such as the March ofDimes. Ask for a speaker to give a presenta-tion about such genetic diseases, including thespecific point mutations, the gene and chro-mosome involved, and the consequences ofthe mutations. TAKS 2 Bio 6C

MEDICINEMEDICINECONNECTIONCONNECTION

English Language Learners

HOH2C

C

H

H

C

OH

CH

HC

H

OHO

HOH2C

C

H

H

C

OH

CH

HC

OH

OHO

Deoxyribose

Ribose

TeachTeach

Species of Amanita including flyagaric (A. muscaria) and pan-thercap (A. pantherina), containthe poisons ibotenic acid andmuscimol. These poisons arerelated and have a similar affecton the body, although muscimolis generally more potent. Thesepoisons affect the nervous sys-tem, and may lead to drowsiness,dizziness, delirium, and hyperac-tivity. Unlike the liver-damagingpoison found in A. phalloides,these poisons are rarely fatal, butcan cause coma and convulsionsin children. TAKS 2 Bio 4B

Real Life

Real LifeDeath cap mushroomsare deadly if eaten. One of the poisons indeath cap mushrooms(Amanita phalloides) istaken up by liver cells,where the poison binds toan RNA polymerase. Thepoison prevents liver cellsfrom making RNA and,thus, from making pro-teins. Liver failure—anddeath—can result.Finding InformationResearch other poisonsfound in Amanitaspp. and determinetheir methodsof action.

Transfer of Information from DNA to RNA The first step in the making of a protein, transcription, takes theinformation found in a gene in the DNA and transfers it to a mol-ecule of RNA. , an enzyme that adds and linkscomplementary RNA nucleotides during transcription, is required.Figure 2 summarizes the steps of transcription.

Step Transcription begins when RNA polymerase binds to thegene’s promoter—a specific sequence of DNA that acts as a“start” signal for transcription.

Step RNA polymerase then unwinds and separates the twostrands of the double helix, exposing the DNA nucleotideson each strand.

Step RNA polymerase adds and then links complementary RNAnucleotides as it “reads” the gene. RNA polymerase movesalong the nucleotides of the DNA strand that has the gene,much like a train moves along on a track. Transcription fol-lows the base-pairing rules for DNA replication except thatin RNA, uracil, rather than thymine, pairs with adenine.

As transcription proceeds, the RNA polymerase eventuallyreaches a “stop” signal in the DNA. This “stop” signal is a sequenceof bases that marks the end of each gene in eukaryotes, or the endof a set of genes in prokaryotes.

RNA polymerase

BIOgraphic

RNAPromotersite on DNA

RNApolymerase

RNA polymerase binds to the gene’s promoter.

RNA polymerase adds complementary RNA nucleotides as it reads the gene.

Transcription: Making RNA

1 The two DNA strands unwind and separate.

2 Complementary RNA nucleo-tides are added.

3

Figure 2

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Teaching TipComparing Transcription andReplication Have students make a Graphic Organizer similar to theone at the bottom of this page toillustrate the differences betweentranscription and DNA replication.

Verbal

Brainstorming Pair each studentwith a partner. Have each pair readthe first page of the section aloud,sharing the reading equally. Thenask students to answer the follow-ing questions:

• What is the main idea in thissection? What passages or wordsled you to this conclusion?

• How will this section be organized?What words or sentences supportyour conclusion?

AuditoryLS

SKILLBUILDER

READINGREADING

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Teach, continuedTeach, continued


ModelingTranscriptionSkills AcquiredAnalyzing, predicting

Answers to Analysis1. Two colors represent the two

different molecules.2. The mRNA sequence would

not be the same as the oneconstructed in the activity.

3. Their second mRNA is differ-ent from the first mRNA.


Graphic Organizer

Use this graphic organizer with Teaching Tip on this page.

Transcription DNA replication

RNA polymerase is used. DNA polymerase is used.

RNA nucleotides are linked. DNA nucleotides are linked.

An RNA molecule is made. A DNA molecule is made.

Only one part of one strand Both DNA strands serve as(a gene) is used as a template. templates.

When the RNA nucleotides are added during transcription, theyare linked together with covalent bonds. As RNA polymerasemoves down the strand, a single strand of RNA grows. BehindRNA polymerase, the two strands of DNA close up by forminghydrogen bonds between complementary bases, re-forming theDNA double helix.

Like DNA replication, transcription uses DNA nucleotides as atemplate for making a new molecule. However, in DNA replication,the new molecule made is DNA. In transcription, the new moleculemade is RNA. In addition, in DNA replication, both strands of DNAserve as templates, whereas in transcription, only part of one ofthe two strands of DNA (a gene) serves as a template.

Transcription in prokaryotic cells occurs in the cytoplasm(because prokaryotic cells have no nucleus); transcription ineukaryotic cells occurs in the nucleus, where the DNA is located.During transcription, many identical RNA molecules are madesimultaneously from a single gene, as shown in Figure 3. The RNAbeing made fans out from the gene to give a “feathery” appear-ance. The long line along the length of the “feather” is the DNAbeing transcribed. The circles along the length are the RNA poly-merase molecules. The “hairs” on the feather are the RNA chainsbeing made.

Figure 3 Multiple copiesof RNA. In eukaryotes, RNApolymerase adds about 60nucleotides per second. There are typically about 100RNA polymerase moleculesper gene.

Modeling TranscriptionYou can use paper and pens to model the process of transcription.

Materials

paper, scissors, pens or pencils (two colors), tape

Procedure

1. Cut a sheet of paperinto 36 squares, each

about 2.5 � 2.5 cm (1 � 1 in.) in size.

2. To make one side of yourDNA model, line up 12 squares in a column.Using one color, randomlylabel each square with one ofthe following letters: A, C, G,or T. Each square represents a DNA nucleotide. Use tape to keep the squares in a column.

3. To make the second side of your DNA model, line up12 squares next to the firstcolumn. Use the same coloryou used in step 2 to labeleach square with the comple-mentary DNA nucleotide.Tape the squares together ina column.

4. Separate the two columns.The remaining 12 squaresrepresent RNA nucleotides.Use a different color to “transcribe” one of the DNA strands.

Analysis

1. Propose a reason for usingdifferent colors for the DNAand RNA “nucleotides.”

2. Predict how a change in thesequence of nucleotides in aDNA molecule would affectthe mRNA transcribed fromthe DNA molecule.

3. Critical ThinkingApplying Information Useyour model to test your pre-diction. Describe your results.

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pp. 210–211

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ActivityTriplet Spelling Write the fourletters A, E, R and T on the board.Ask students to write as manymeaningful three-letter words asthey can using any three of the fourletters. (Answers may include art,are, rat, tar, tea, eat, ate, ear.)Continue by asking students toform words that use the lettersmore than once. (New answers mayinclude tee, and tat.) Studentsshould see that many three-lettercombinations can be produced.Compare this exercise to the tablein Figure 4. Ask students how thetwo compare. (In the exercise, lettersare combined to form words, but in the table the letters stand for combinations of nucleotides.)

Math Skills Tell students they canpredict the number of possiblecombinations if they know howmany items there are to choosefrom and how many items will bein a set. For example, there are 4possible bases in an mRNA codon,and three bases per codon. Thatpredicts a possible 4 � 4 � 4 � 64possibilities. Have students confirmthis in the table in Figure 4. Askstudent to predict how many possi-ble codons would exist if therewere only 2 bases per codon?(4 � 4 � 16) How many codonswould exist if there were only threepossible bases, and 3 bases percodon? (3 � 3 � 3 � 27) Logical

Interactive Reading AssignChapter 10 of the Holt BiologyGuided Audio CD Program to helpsutdents achieve greater success in reading thechapter.

SKILLBUILDER

READINGREADING

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GENERALBUILDERSKILL

GENERAL


MISCONCEPTION ALERT

Translation When given an mRNA strandto translate, many students mistakenly usethe anticodons, instead of the codons, todetermine the amino acid sequence. Pointout that the genetic code is based on thecodons found on the mRNA and not on theanticodons of tRNA. The codon sequence isthe genetic code (DNA language) rewritten ortranscribed in RNA language. TAKS 2 Bio 6B

(grade 11 only)

The prokaryotic ribosome is smaller than theeukaryotic ribosome. Its protein and RNAcontent are also dissimilar. These differencesallow selective antibiotics, such as tetracy-cline, to bind to the prokaryotic ribosomesand interfere with prokaryotic proteinsynthesis. These antibiotics can be safelygiven to humans because they do not affectthe ribosomes and protein synthesis ofeukaryotic cells. TAKS 2 Bio 4B

REAL WORLDREAL WORLDCONNECTIONCONNECTION



Codons in mRNA

First Second base Third base U C A G base

UUUPhenylalanine

UCU UAUTyrosine

UGUCysteine

UU UUC UCC

SerineUAC UGC C

UUALeucine

UCA UAAStop

UGA –Stop AUUG UCG UAG UGG–Tryptophan G

CUU CCU CAUHistidine

CGU UC CUC

LeucineCCC

ProlineCAC CGC

ArginineC

CUA CCA CAAGlutamine

CGA ACUG CCG CAG CGG G

AUU ACU AAUAsparagine

AGUSerine

UA AUC Isoleucine ACC

ThreonineAAC AGC C

AUA ACA AAALysine

AGAArginine

AAUG –Start ACG AAG AGG G

GUU GCU GAU Aspartic GGU UG GUC

ValineGCC

AlanineGAC Acid GGC

GlycineC

GUA GCA GAA Glutamic GGA AGUG GCG GAG Acid GGG G

The amino acid coded for by a specific mRNA codoncan be determined by following the three steps below.

1. Find the firstbase of themRNA codonalong the leftside of thetable.

2. Follow that rowto the rightuntil you arebeneath thesecond base of the codon.

3. Move up ordown in thatsection untilyou are even,on the rightside of thechart, with thethird base ofthe codon.

The Genetic Code: Three-Nucleotide “Words”Different types of RNA are made during transcription, dependingon the gene being expressed. When a cell needs a particular pro-tein, it is messenger RNA that is made. (mRNA) isa form of RNA that carries the instructions for making a proteinfrom a gene and delivers it to the site of translation. The informa-tion is translated from the language of RNA—nucleotides—to thelanguage of proteins—amino acids. The RNA instructions arewritten as a series of three-nucleotide sequences on the mRNAcalled (KOH dahnz). Each codon along the mRNA strandcorresponds to an amino acid or signifies a start or stop signal fortranslation.

In 1961, Marshall Nirenberg, an American biochemist, deci-phered the first codon by making artificial mRNA that containedonly the base uracil (U). The mRNA was translated into a proteinmade up entirely of phenylalanine amino-acid subunits.Nirenberg concluded that the codon UUU is the instruction forthe amino acid phenylalanine. Later, scientists deciphered theother codons. Figure 4 shows the —the amino acidsand “start” and “stop” signals that are coded for by each of thepossible 64 mRNA codons.

genetic code

codons

Messenger RNA

Figure 4 Interpreting the genetic code

www.scilinks.orgTopic: Genetic CodeKeyword: HX4089

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Teaching TipRibozymes A substance calledpeptidyl transferase catalyzes theformation of a peptide bond thatjoins the polypeptide chain fromthe P site to the amino acid at theA site. Evidence suggests that pep-tidyl transferase is not a protein, asmight be expected, but an rRNAmolecule that acts as an enzyme.RNA catalysts are known asribozymes. Ask students whatimpact this evidence has on previ-ously held beliefs about biologicalcatalysts. (All biological catalystswere once thought to be protein innature; ribozymes are nucleic acids.)

Using the Figure

Work with students tohelp them summarize the eventsthat take place during translationshown in Figure 5. Point out thecomplementary nature of codonson the mRNA and the anticodonson the tRNA, as shown in STEP1. Refer students to the table inFigure 4 to make the connectionbetween the amino acid thetRNA is carrying and the codonit binds to. Be sure studentsunderstand the function of eachof the following: mRNA (carriescode for making protein), tRNA(carries specific amino acids to site oftranslation), and ribosome(coordinates protein assembly). Then,ask students to close their booksand write a summary oftranslation. VisualTAKS 2 Bio 4B; 6B (grade 11 only)

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In 2002, as the human genome project wasnearing completion of its first goals, scientistshad sequenced nearly all 3 billion base pairs.These base pairs include about 30,000 genes.


did you know?E Sites In addition to the A and P sites,researchers have discovered a third tRNA bind-ing site on the ribosome—the E (exit) site. Whena tRNA molecule detaches, as shown in Step 4of Figure 5, the tRNA first moves from the Psite to the E site on the ribosome. The tRNA isthen released from the ribosome. TAKS 2Bio 6B (grade 11 only)

BIOgraphic BIO

graphicTranslation: Assembling Proteins

Amino acids are assembled from information encoded in mRNA.

Nuclear envelope

Nuclear pore

mRNA

Amino acid methionine

(Met)

tRNAAmino acid

Psite

Asite

Ribosome

Met

The ribosomal subunits, the mRNA, and the tRNA carrying methionine bind together.

1 The tRNA carrying the amino acid specified by the codon in the A site arrives.

2 A peptide bond forms between adjacent amino acids.

3

RNA’s Roles in TranslationTranslation takes place in the cytoplasm. Here transfer RNA mol-ecules and ribosomes help in the synthesis of proteins.

(tRNA) molecules are single strands of RNA thattemporarily carry a specific amino acid on one end. Each tRNA isfolded into a compact shape and has an anticodon (an tee KOHdahn). An is a three-nucleotide sequence on a tRNAthat is complementary to an mRNA codon. As shown in Figure 5,the amino acid that a tRNA molecule carries corresponds to a par-ticular mRNA codon.

Ribosomes, shown in Figure 5, are composed of both proteinsand ribosomal RNA (rRNA). molecules are RNAmolecules that are part of the structure of ribosomes. A cell’s cyto-plasm contains thousands of ribosomes. Each ribosome tem-porarily holds one mRNA and two tRNA molecules. Figure 5summarizes the process of translation:

Step Translation begins when the mRNA leaves the nucleus andenters the cytoplasm. The mRNA, the two ribosomal sub-units, and a tRNA carrying the amino acid methionine(muh THIE uh neen) together form a functional ribosome.The mRNA “start” codon AUG, which signals the beginningof a protein chain, is oriented in a region of the ribosomecalled the P site, where the tRNA molecule carrying methio-nine can bind to the start codon.

Ribosomal RNA

anticodon

Transfer RNA

Figure 5

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Student EditionTAKS Obj 2 Bio 4B TAKS Obj 2 Bio 6B (grade 11)TEKS Bio 4B, 6B

Teacher EditionTAKS Obj 2 Bio 4B, 6A, 6BTAKS Obj 3 Bio 4CTAKS Obj 4 IPC 8ATEKS Bio 3F, 4B, 4C, 6A, 6B, 8A, 9A

pp. 212–213

Group ActivityRetroviruses Retroviruses containRNA rather than DNA as theirnucleic acid, and they use RNA asa template to make DNA. To dothis, an enzyme called reversetranscriptase is used. HIV is anexample of one such virus. Severaldrugs have been developed to slowthe replication of the virus. Two ofthese drugs are lamivudine andzidovudine. Have students work inteams of three or four to researchhow these drugs function, write abrief report on their findings, andmake simple models or drawingscomparing DNA to RNA. (Thesedrugs inhibit the action of reversetranscriptase. In retroviruses, DNA is made from viral RNA. This DNA is incorporated into the hostDNA, which is then transcribed, result-ing in the viral mRNA being translatedto make specific viral proteins.)

Verbal

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Co-op LearningLS


Decoding theGenetic CodeSkills AcquiredRecognizing patterns,interpreting information

Teacher’s NotesUse Figure 5 to review theprocess of translation. Makesure students understand theterms codon and anticodon.Point out that because of spacelimitations, the start and stopcodons are not included on themRNA strand that is shown.

Answers to Analysis1. Serine-arginine-glutamic acid-

phenylalanine-serine2. AGA, GCA, CUU, AAA, AGG3. AGAGCACTTAAAAGG4. TCTCGTGAATTTTCC

010001011001110101000100100010011100100100010000010100100111010101001000101010010010

StrategiesStrategiesINCLUSIONINCLUSION

Ask students to research the life of BarbaraMcClintock. The students should findinformation about her life and her work.Additionally, have students give their opinionof how it would feel and how they wouldreact if they made a major scientific discoverythat the scientific world denounced andwould not recognize for twenty years. Theycan report their findings and opinions in atape-recorded message to the class. Bio 3F

• Attention Deficit Disorder

• Learning Disabilities

Met

Met

Met

Met

Growingprotein chain

Newlymade

protein

The tRNA in the P site detaches and leaves its amino acid behind.

4 The tRNA in the A site moves to the P site. The tRNA carrying the amino acid specified by the codon in the A site arrives.

5 A peptide bond is formed. The tRNA in the P site detaches and leaves its amino acid behind.

6 The process is repeated until a stop codon is reached. The ribosome complex falls apart. The newly made protein is released.

7

Step The codon in the area of the ribosome called the A site isready to receive the next tRNA. A tRNA molecule with thecomplementary anticodon arrives and binds to the codon.The tRNA is carrying its specific amino acids.

Step Now both the A site and the P site are holding tRNA mol-ecules, each carrying a specific amino acid. Enzymes thenhelp form a peptide bond between the adjacent amino acids.

Step Afterward, the tRNA in the P site detaches, leaves behind itsamino acid, and moves away from the ribosome.

Step The tRNA (with its protein chain) in the A site moves overto fill the empty P site. Because the anticodon remainsattached to the codon, the tRNA molecule and mRNA mol-ecule move as a unit. As a result, a new codon is present inthe A site, ready to receive the next tRNA and its aminoacid. An amino acid is carried to the A site by a tRNA andthen bonded to the growing protein chain.

Step The tRNA in the P site detaches and leaves its amino acid.

Step Steps 2 through 6 are repeated until a stop codon is reached.A stop codon is one of three codons (UAG, UAA, or UGA) forwhich there is no tRNA molecule with a complementaryanticodon. Because there is no tRNA to fit into the empty A site in the ribosome, protein synthesis stops. The newlymade protein is released into the cell.

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IPC Benchmark Fact

Point out that the metabolic processes by which E. coli absorb and break down lactose is a form ofdigestion whereby food—“milk sugar” in thisinstance—is chemically changed into simpler com-pounds that our bodies can then use for energy ormaking other structures needed by cells. Ask stu-dents to name the enzyme involved in the chemicaldigestion of lactose and to identify the portion of itsname that indicates it is an enzyme. TAKS 4 IPC 8A

ReteachingWrite the following on eight differ-ent pieces of paper: transcription,translation, DNA, RNA, mRNA,tRNA, codon, and anticodon. Putthese pieces of paper in a smallcontainer. Write the question Howare they linked? on the board.Have a student pick two pieces ofpaper from the container. Show theterms to the class and give them acouple of minutes to write downthe answer. Return the papers tothe container and repeat as long astime will allow.

QuizTrue or False:

1. RNA is similar in structure toDNA, except it contains thesugar uracil rather than thymine.(False. RNA does contain uracil,but uracil is a base.)

2.The anticodon is the comple-mentary sequence of the codon.(True. Codons are mRNA, whileanticodons are tRNA)

3. The codons are the same formost organisms. (True. Genesmay differ, but the codons withinthe genes are the same.)

AlternativeAssessmentHave students create a colorfulposter that compares and illustratesthe functions of mRNA, tRNA, and rRNA. Bio 9A

GENERAL

GENERAL

Bio 9A

CloseClose

Answers to Section Review

1. RNA is single stranded; DNA is doublestranded. RNA contains the sugar ribose; DNAcontains the sugar deoxyribose. RNA containsthe bases A, G, C, U; DNA contains the basesA, G, C, T.

2. RNA polymerase binds to the promoter,unwinds and separates the DNA strands, thenadds and links complementary RNAnucleotides as it “reads” the gene.

3. proline4. mRNA carries the instructions for making a

protein; tRNA temporarily carries a specificamino to the site of translation; rRNA is partof the ribosome. TAKS 2 Bio 6B (grade 11 only);

Bio 9A

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TAKS 2 Bio 6A

5. Transcription means “writing out;” instruc-tions on the gene are written out as mRNA.Translation means, “to put into words of a dif-ferent language;” the instructions for making aprotein are translated from the language ofnucleic acids to amino acids.

6. A. Incorrect. Codons are based on sets of threenucleotides, not four. B. Correct. There arefour codons, which consist of three nucleotideseach. C. Incorrect. Codons are not based ontwo-nucleotide segments. D. Incorrect. Codonsare not based on a single nucleotide. TAKS 2 Bio 6B

(grade 11 only)



As the mRNA moves across the ribosome, another ribosome canfind the AUG codon on the same mRNA and begin making a secondcopy of the same protein. In this way many copies of the same proteinare made from a single mRNA molecule.

With few exceptions, the genetic code is the same in all organ-isms. For example, the codon GUC codes for the amino acid valinein bacteria, in eagles, in plants, and in your own cells. For this rea-son, the genetic code is often described as being nearly universal. Itappears that all life-forms have a common evolutionary ancestorwith a single genetic code. Some exceptions include the ways cellorganelles that contain DNA (such as mitochondria and chloro-plasts) and a few microscopic protists read “stop” codons.

Analysis

1. Determine the sequence ofamino acids that will resultfrom the translation of thesegment of mRNA above.

2. Determine the anticodon ofeach tRNA molecule that willbind to this mRNA segment.

3. Critical ThinkingRecognizing PatternsDetermine the sequence ofnucleotides in the segment ofDNA from which the mRNAstrand above was transcribed.

4. Critical ThinkingRecognizing PatternsDetermine the sequence ofnucleotides in the segment ofDNA that is complementary tothe DNA segment describedin item 3.

Decoding the Genetic CodeBackground

Keratin is one of the proteins in hair. The gene for keratin istranscribed and translated by certain skin cells. The series ofletters below represents the sequence of nucleotides in a por-tion of an mRNA molecule transcribed from the gene for ker-atin. This mRNA strand and the genetic code in Figure 4 can beused to determine some of the amino acids in keratin.

010001011001110101000100100010011100100100010000010100100111010101001000101010010010

A AU U U U U U UC C C CG G

Section 1 Review

Distinguish two differences between RNAstructure and DNA structure. 6A

Explain how RNA is made during transcription.

Interpret the genetic code to determine theamino acid coded for by the codon CCU. 6A

Compare the roles of the three different types ofRNA during translation. 6B 9A

Critical Thinking Justifying ConclusionsEvaluate the following statement: The termtranscription is appropriate for describing theproduction of RNA, and the term translation isappropriate for describing the synthesis of proteins.

What is the maximum numberof amino acids that could be coded for by a sectionof mRNA with the sequence GUUCAGAACUGU? A 3 C 6B 4 D 12

TAKS Test PrepTAKS Test Prep

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6A

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Teacher Edition TAKS Obj 2 Bio 6CTEKS Bio 6C, 9A

pp. 214–215

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Section 2

OverviewBefore beginning this sectionreview with your students theobjectives listed in the StudentEdition. In this section, studentswill learn that cells express only asmall percentage of the genes theycontain at any given moment. Themethods of gene regulation—orhow genes are turned “on” and“off” are examined for bothprokaryotes and eukaryotes.Students will also learn about thestructure of eukaryotic genes, andthe ways that mutations can alterthe function of genes.

Challenge students by writing thefollowing question on the board:Why has protein synthesis researchbeen focused on prokaryotes ratherthan eukaryotes? Have themanswer the question in theirnotebooks, and then discuss.(Prokaryotes are easier to grow andsimpler to study. Like eukaryotes,prokaryotes have DNA, genes, andcodons, but prokaryotes have muchless DNA, and their DNA is circularrather than linear.) Verbal

DemonstrationHold up a copy of the student edi-tion textbook. Point out to studentsthat their textbook is about 1,000pages long. Ask students to imaginethe textbook as 1,000 pages ofgenetic information found withineach of their cells. A typical cell atany given time is only using 3–5percent of this information, or30 to 50 pages. Flip through 30–50pages to illustrate this—about onechapter’s worth.

MotivateMotivate

Bio 9ALS

Bellringer

TAKS 2 Bio 6C

FocusFocus


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

GENERAL

GENERAL


Transparencies

TT BellringerTT Controlling Transcription in

EukaryotesTT Major Types of Mutations


• Reading Organizers• Reading Strategies• Supplemental Reading Guide

A Feeling for the Organism

Planner CD-ROM

Operator1

2 3

Promoter

Genes Involved in lactose useRNA polymerase

Lactose absent—the lac operon is off.

Lactose present—the lac operon is on.

Repressor protein

Operator1

2 3

Promoter

Transcription proceedsLactose bound to repressor protein

Lactose

Protein Synthesis in Prokaryotes Although prokaryotic organisms, such as bacteria, might seem simplebecause of their small size, prokaryotic cells typically have about2,000 genes. The human genome, which is the largest genomesequenced to date, has about 30,000 genes. Not all of the genes, how-ever, are transcribed and translated all of the time; this would wastethe cell’s energy and materials. Both prokaryotic and eukaryotic cellsare able to regulate which genes are expressed and which are not,depending on the cell’s needs.

An example of gene regulation that is well understood in prokary-otes is found in the bacterium Escherichia coli. When you eat or drinka dairy product, the disaccharide lactose (“milk sugar”) reaches theintestinal tract and becomes available to the E. coli living there. Thebacteria can absorb the lactose and break it down for energy or formaking other compounds. In E. coli, recognizing, consuming, andbreaking down lactose into its two components, glucose and galac-tose, requires three different enzymes, each of which is coded for bya different gene.

As shown in Figure 6, the three lactose-metabolizing genes arelocated next to each other and are controlled by the same promotersite. There is an on-off switch that “turns on” (transcribes and thentranslates) the three genes when lactose is available and “turns off”the genes when lactose is not available.

Gene Regulation and Structure

Section 2

Objectives● Describe how the lac

operon is turned on or off.

● Summarize the role oftranscription factors inregulating eukaryotic geneexpression.

● Describe how eukaryotic genes are organized.

● Evaluate three ways that point mutations can alter genetic material.

Key Terms

operatoroperonlac operonrepressorintronexonpoint mutation

The lac operon allows a bacterium to build the proteins needed for lactosemetabolism only when lactose is present.

Figure 6 Turning prokaryotic genes on and off

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4A 6A

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4B 6A 6C

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DemonstrationCut shapes from colored paper torepresent the components of the lacoperon as shown in Figure 6. Theseshapes should be large enough forstudents to see from their seats.Secure these shapes to the boardand use them to demonstrate theprocess by which genes are turnedon and off in the lac operon.Emphasize the role of feedback systems in gene regulation.

VisualLS

GENERAL

TeachTeach


Jumping GenesTeaching Strategies• Help make transposons rele-

vant to students’ lives bypointing out that one type oftransposon has been identifiedas a cause of leukemia. Inhumans, a piece of chromo-some 22 breaks off and bindsto chromosome 9.

• Tell students that transposase,an enzyme encoded by trans-posons, is responsible fortransposition. The movementof transposase to differentsites of the genome is randomand rare.

Discussion• What effect do transposons

have on other genes?(Transposons can inactivate thegenes they jump into.)

CulturalAwarenessCulturalAwareness

The lac Operon Two French scientists,Francois Jacob and Jacques Monod, wereresponsible for the discovery of operons.They both joined the staff of InstitutPasteur, where they collaborated on geneticresearch. By 1960, these two men were ableto determine how genes influence the mak-ing or degrading of certain substances—asystem referred to as an operon. They wereawarded a Nobel Prize in 1965. Bio 3F


Many adults lack sufficient lactase enzymesto digest the lactose found naturally in dairyproducts. This can lead to gas, bloating, anddiarrhea after consuming products containinglactose. Lactase enzymes are available as adietary supplement, and in some products,lactase is added before consumption to break down the lactose sugar into moredigestible sugars. Bio 9A

REAL WORLDREAL WORLDCONNECTIONCONNECTION

Jumping Genes

The spotted and streaked patterns seen in Indian corn result from genes that have movedfrom one chromosomal location to another. Such genes are called transposons (trans POH zahns). When a transposon jumps to a new location, it often inactivates a gene orcauses mutations. In Indian corn, some pigment genes are not expressed in some cells because they have been disrupted byjumping genes.

The Discovery of TransposonsIn the 1950s, the geneticist Barbara McClintock discovered transposons whilestudying corn. Most scientists rejected her ideas for more than 20 years. The idea that genes could change locations on the chromo-some contradicted the prevailing view that genes and chromosomes are stable parts of the cell. Over time, additional researchsupported her hypothesis, and her modelgradually gained acceptance. In 1983, McClintock received a Nobel Prize for her discoveries involving transposons.

Importance of TransposonsAll organisms,including humans,appear to havetransposons.Transposons proba-bly play a role inspreading genes forantibiotic resistanceamong bacteria.Transposons thataffect flower colorin morning gloryflowers have beenfound. Transposonsmay also havemedical applications,such as helping scien-tists discover how white blood cells make antibod-ies and what causes cancer.

Although the movement of transposons is veryrare, transposons are important because they cancause mutations and bring together differentcombinations of genes. The transfer of these mobilegenes could be a powerful mechanism in evolutionand could help solve certain mysteries about evolu-tion, such as how larger organisms developed fromsingle cells and how new species arise.

FurtherExploring Further

The piece of DNA that overlaps the promoter site and serves asthe on-off switch is called an . Because of its position, theoperator is able to control RNA polymerase’s access to the three lac-tose-metabolizing genes.

In bacteria, a group of genes that code for enzymes involved in thesame function, their promoter site, and the operator that controlsthem all function together as an (AHP uhr ahn). The operonthat controls the metabolism of lactose is called the andis shown in Figure 6.

What determines whether the lac operon is in the “on” or “off”mode? When there is no lactose in the bacterial cell, a repressor turnsthe operon off. A is a protein that binds to an operator andphysically blocks RNA polymerase from binding to a promoter site.The blocking of RNA polymerase consequently stops the transcrip-tion of the genes in the operon, as shown in Figure 6.

When lactose is present, the lactose binds to the repressor andchanges the shape of the repressor. The change in shape causesthe repressor to fall off of the operator, as shown in Figure 6. Nowthe bacterial cell can begin transcribing the genes that code for the lactose-metabolizing enzymes. By producing the enzymes onlywhen the nutrient is available, the bacterium saves energy.

repressor

lac operonoperon

operator

Barbara McClintock

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Bio 3E

TAKS 2 Bio 6A;Bio 3F

Student EditionTAKS Obj 2 Bio 4B TAKS Obj 2 Bio 6A TAKS Obj 2 Bio 6B (grade 11)TAKS Obj 2 Bio 6CTEKS Bio 3F, 4B, 6A, 6B, 6C, 9A

Teacher EditionTAKS Obj 2 Bio 6A, 6BTAKS Obj 2 Bio 6BTEKS Bio 3D, 3E, 3F, 6A, 6B, 9A

pp. 216–217

Trends in BiotechnologyDesigning Proteins Current computing sys-tems can simulate the folding of a protein into itscompact form, but it takes 40 months of proces-sor time to run such a simulation. Advances interascale computing will allow one trillion opera-tions per second. Instead of 40 months, the simu-lation would take just one day. Such technologywould allow scientists to custom design proteinsfor medical treatments and other applications.Bio 9A

Teaching TipAdult versus Fetal Genes Fetalhemoglobin, which differs some-what from adult hemoglobin, has ahigher affinity for oxygen. A shorttime after birth, fetal hemoglobin isreplaced by adult hemoglobin. Wedo not know what mechanismstops the production of fetal hemo-globin and begins the production ofadult hemoglobin. Apparently, thegenes for fetal hemoglobin areturned off, and those for adulthemoglobin are turned on just after birth.

Teaching TipUnexpressed Genes A typicalhuman cell only expresses 3–5percent of its genes at any giventime. Also, not all mRNA is trans-lated once it is made. The egg cellsof many organisms synthesize andstore large amounts of mRNA molecules that are translated afterfertilization. Bio 9A

Bio 9A

GENERAL


CareerCareerMolecular Geneticist Molecular geneticsdeals with the molecular nature of genes andtheir role in the function and development of anorganism. Molecular geneticists with a B.S.degree can work as laboratory technicians. With advanced degrees, molecular geneticistscan design and supervise research projects.Molecular geneticists work for universities, government agencies, and agricultural, pharmaceutical, and biotechnological firms.Bio 3D

Enhancer

Activator RNA polymerase Transcriptionfactor

Enhancer

Promoter

Promoter

Transcription begins

Coding region of gene

Coding region of gene

DNA

Protein Synthesis in Eukaryotes Eukaryotic cells contain much more DNA than prokaryotic cells do.Like prokaryotic cells, eukaryotic cells must continually turn certaingenes on and off in response to signals from their environment.Operons have not been found often in eukaryotic cells. Instead, geneswith related functions are often scattered on different chromosomes.

Because a nuclear envelope physically separates transcriptionfrom translation in a eukaryotic cell, more opportunities exist forregulating gene expression. For example, gene regulation can occurbefore, during, and after transcription. Gene regulation can alsooccur after mRNA leaves the nucleus or after translation, when theprotein is functional.

Controlling the Onset of TranscriptionMost gene regulation in eukaryotes controls the onset of transcrip-tion—when RNA polymerase binds to a gene. Like prokaryotes,eukaryotic cells use regulatory proteins. But many more proteins areinvolved in eukaryotes, and the interactions are more complex. Theseregulatory proteins in eukaryotes are called transcription factors.

As shown in Figure 7, transcription factors help arrange RNApolymerases in the correct position on the promoter. A gene can beinfluenced by many different transcription factors.

An enhancer is a sequence of DNA that can be bound by a tran-scription factor. Enhancers typically are located thousands ofnucleotide bases away from the promoter. A loop in the DNA may bringthe enhancer and its attached transcription factor (called an activator)into contact with the transcription factors and RNA polymerase at the promoter. As shown in Figure 7, transcription factors bound to enhancers can activate transcription factors bound to promoters.

Organizing InformationMake a table to organizeinformation about the regu-lation of protein synthesis.Across the top write theheadings Prokaryotes andEukaryotes. Along the sideswrite Protein(s) that regu-late(s) the genes andDetails of regulation. Addinformation to the table asyou read Section 2.

Transcription factors bind to the enhancer and to the RNA polymerase. Thebinding activates transcription factors bound to the promoter.

Figure 7 Controlling transcription in eukaryotes

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IPC BenchmarkMini Lesson

Biology Skills TAKS 2 Bio 6B (grade 11only) Explain replication, transcription,and translation using models. ActivityHave students work in small groups anduse colored beads, or snap-togetherbuilding blocks to simulate replication,transcription, translation and proteinsynthesis. Check their understanding ofeach process by having each studentdemonstrate one of the processes tothe others in their group.

Teaching TipIntrons and Exons Tell studentsthat two scientists independentlyfound evidence that introns andexons existed. These two scientists,Richard Roberts and Phillip Sharp,shared a Nobel Prize in 1993 forthis discovery. We now know thatthe cutting and pasting of intronsand exons is carried out by acomplex of proteins and RNAmolecules together called snRNP(pronounced “snurps”), or smallriboncucleoproteins. Bio 3F


DNA. In a eukaryotic cell, mRNA can persistfor hours, and sometimes for days or weeks.Compare this to two minutes in a prokaryoticcell. The function of the poly A tail isunknown, although it is though to aid in theexport of mRNA from the nucleus and toprevent degradation in the cytoplasm. Bio 9A


Modeling Intronsand ExonsSkills AcquiredApplying information, predicting outcomes

Teacher’s NotesReview the terms intron andexon with the students. Ask stu-dents to explain the graphicbefore beginning the lab.

Answers to Analysis1. The strip with the letters

apprialyjoed represents theintrons. The strip with the letters that spell protein represent the exons.

2. Answers will vary. Because thefunction of a protein is ulti-mately a result of its aminoacid sequence, a protein withadditional amino acids willmost likely not function.

did you know?Eukaryotic mRNA At one end of the mRNAmolecule, a cap consisting of a nucleotide called7-methylguanylate attaches to the molecule. Atthe other end of mRNA, a poly A tail, consist-ing of many adenines, attaches. Scientists thinkthat the cap prevents the mRNA from degrada-tion, making it more stable than prokaryotic

Intervening DNA in Eukaryotic GenesWhile it is tempting to think of a gene as an unbroken stretch ofnucleotides that code for a protein, this simple arrangement is usuallyfound only in prokaryotes. In eukaryotes, many genes are interruptedby (IN trahnz)—long segments of nucleotides that have nocoding information. (EK sahnz) are the portions of a gene thatare translated (expressed) into proteins. After a eukaryotic gene istranscribed, the introns in the resulting mRNA are cut out by complexassemblies of RNA and protein called spliceosomes. The exons thatremain are “stitched” back together by the spliceosome to form asmaller mRNA molecule that is then translated.

Many biologists think this organization of genes adds evolution-ary flexibility. Each exon encodes a different part of a protein. Byhaving introns and exons, cells can occasionally shuffle exonsbetween genes and make new genes. The thousands of proteins thatoccur in human cells appear to have arisen as combinations of onlya few thousand exons. Some genes in your cells exist in multiplecopies, in clusters of as few as three or as many as several hundred.For example, your cells each contain 12 different hemoglobin genes,all of which arose as duplicates of one ancestral hemoglobin gene.

Exonsintrons

The “int” in the word introncomes from the “int” in theword intervening. The “ex”in the word exon comesfrom the “ex” in the wordexpressed.

Modeling Introns and ExonsYou can use masking tape to represent introns and exons.

Materials

masking tape, pens or pencils (two colors), metric ruler, scissors

Procedure

1. Place a 15–20 cm strip ofmasking tape on your desk.The tape represents a gene.

2. Use two colors to write thewords appropriately joined onthe tape exactly as shown inthe diagram below. Space theletters so that they take upthe entire length of the strip of tape. The segments in onecolor represent introns; thosein the other color representexons.

3. Lift the tape. Workingfrom left to right, cut

apart the groups of letterswritten in the same color.Stick the pieces of tape toyour desk as you cut them,making two strips accordingto color and joining the piecesin their original order.

Analysis

1. Determine from the result-ing two strips which strip ismade of “introns” and whichis made of “exons.”

2. Critical ThinkingPredicting OutcomesPredict what might happen toa protein if an intron were notremoved.

Transcription

Intronsremoved

mRNA leavesnucleus

Translation

IntronExon

mRNA(exons spliced together)

mRNA

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Bio 3E; TAKS 2 Bio6A, 6B (grade 11 only)

Student EditionTAKS Obj 2 Bio 4B TAKS Obj 2 Bio 6ATAKS Obj 2 Bio 6C TEKS Bio 3E, 4B, 6A, 6C

Teacher EditionTAKS Obj 2 Bio 6A, 6B, 6CTEKS Bio 3E, 3F, 6A, 6B, 6C, 9A

pp. 218–219

TAKS 2

Teaching TipMutagens Many environmentalfactors (such as X rays and gammarays) and chemicals can causemutations. Have pairs of studentsresearch a specific mutagen. Theirreport should include the type ofmutation caused and the ill effectsof the mutation on the human body.

TAKS 2 Bio 6CCo-op Learning


HISTORYHISTORYCONNECTIONCONNECTION

In 1986, there was an explosion of a nuclearreactor at the Chernobyl power plant, in theformer Soviet Union (now the Ukraine).Scientists predicted that the residents of thesurrounding towns would develop high ratesof thyroid cancer and leukemia as a result ofradiation exposure. Have students read arti-cles about the effects of the explosion andthen discuss their findings in brief reports thatlink the elevated incidence of cancer to radia-tion damage of DNA. TAKS 2 Bio 6C

did you know?Mitochondrial DNA Mothers and their off-spring have identical mitochondrial DNAbecause sperm mitochondria are destroyed inthe developing zygote. Mutations rarely occurin mitochondrial DNA. Families can be tracedby this DNA because it remains essentiallyunchanged from generation to generation.Bio 9A

Gene SequencingTeaching Strategies• Tell students about the

Human Genome Project—collaboration between manyscientists around the world.Now that all 3 billionnucleotides of the genomehave been sequenced, scien-tists are focused on locatingand identifying the manygenes and the roles they play.

• Tell students that DNAsequencing is made possiblewith the help of enzymes(restriction enzymes), whichcut DNA into pieces foridentification.

DiscussionDNA sequences are collected in computer databases. Whatbenefit does this provide toresearchers? (Long sequencescan be scanned to find particulargenes or promoters.)

GENERAL

The Genetics of MissingTeethTeaching Strategies• Tell students that people

with hypodontia have six orfewer missing permanentteeth. When more than sixpermanent teeth are missing,the condition is calledoligodontia.

• Tell students that scientistsare trying to identify thegenes responsible forhypodontia so that they canpotentially screen for thiscondition in the future anddevelop new forms of treat-ment for the condition.

• Tell students that researchersuse computers to scan andscreen DNA sequences. Theuse of computers enablesresearchers to quickly identifydifferences in DNA sequencesfrom different people.

DiscussionHypodontia is caused by twodifferent types of mutations.How is it possible for two dif-ferent mutations to produce thesame condition? (Both mutationsaffect the correct coding of a pro-tein involved in tooth formation.The insertion mutation leads to asmaller-than-normal protein, thesubstitution mutation leads to anon-functional protein.)

TAKS 2 Bio 6C

GENERAL

Mutations Although changes in an organism’s hereditary information are rel-atively rare, they can occur. As you learned in Chapter 6, a changein the DNA of a gene is called a mutation. Mutations in gametescan be passed on to offspring of the affected individual, but muta-tions in body cells affect only the individual in which they occur.

Mutations that move an entire gene to a new location are calledgene rearrangements. Changes in a gene’s position often disrupt thegene’s function because the gene is exposed to new regulatory con-trols in its new location—like what would happen if you moved toFrance and couldn’t speak French. Two types of gene rearrange-ments are shown in Figure 8. Genes sometimes move as part of atransposon. That is, the genes are carried by the moving transposonlike fleas on a dog. Other times, the portion of the chromosomecontaining a gene may be rearranged during meiosis.

Mutations that change a gene are called gene alterations. Genealterations such as those shown in Figure 8 usually result in theplacement of the wrong amino acid during protein assembly. Thiserror can disrupt the protein’s function. In a , a sin-gle nucleotide changes. In an insertion mutation, a sizable lengthof DNA is inserted into a gene. Insertions often result whenmobile segments of DNA, called transposons, move randomlyfrom one position to another on chromosomes. Transposonsmake up 45 percent of the human genome. In a deletion mutation,segments of a gene are lost, often during meiosis.

point mutation

No Mutation

A B C

A C

A B C

A B B C

Gene Alterations

Point mutation

Insertion

Deletion

Gene RearrangementsTransposition

A C

B

B

Chromosomal rearrangement

A B

B

C

The substitution, addition, or removal of one or morenucleotides is called a gene alteration. If the mutationchanges the original position of a gene of the chromo-some, the gene may not function normally.

Figure 8 Major types of mutations

www.scilinks.orgTopic: Genetic Disorders

Research in TexasKeyword: HXX4008

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ReteachingDivide the class into groups ofthree. Assign each group one of thefollowing topics: lac operon,eukaryotic gene expression, andmutations. Have each group pre-pare a 5-minute presentation thatsummarizes their topic. Each groupshould also prepare a four-questionquiz. Combine the quizzes, andhave the class answer them in writ-ing after the presentations.

Quiz1. In prokaryotes, gene expression

is regulated by ________.(operons)

2.Mutations that change one or afew nucleotides are called________ mutations. (point)

3. True or false: Introns are theparts of a gene that are trans-lated. (False. Introns are segmentsthat are cut out of a gene.)

AlternativeAssessmentForm groups with six to eight stu-dents per group. Have students useeach other to physically demonstratethe operation of the lac operon andto walk through the control of tran-scription in eukaryotes. Each studentshould be able to explain his or herrole in the processes. TAKS 2 Bio 6B

(grade 11 only)

GENERAL

GENERAL

TAKS 2 Bio 6C

CloseClose

Answers to Section Review

1. Lactose binds to the repressor, which changes theshape of the repressor and causes the repressor tofall off of the operator. RNA polymerase is ableto bind, allowing transcription.

2. Transcription factors are regulatory proteins.Some help to arrange RNA polymerase in thecorrect position on the promoter, while others,called activators, bind to an enhancer (a seg-ment of DNA). Transcription begins when theactivator bound to the enhancer comes in con-tact with the transcription factor and RNApolymerase at the promoter.

3. Exons are portions of a eukaryotic gene thatare translated into proteins. Introns contain no

TAKS 2 Bio 4B; 6B(grade 11 only)

TAKS 2 Bio 6A, 6B(grade 11 only)

coding information and are cut out beforetranslation occurs.

4. A frameshift mutation causes a disruption inthe triplet groupings, which results in theincorrect sequence of amino acids. A base-pairsubstitution may or may not change the aminoacid coded for in the triplet.

5. A. Incorrect. A deletion drops anucleotide. B. Incorrect. An insertion adds anucleotide. C. Correct. A substitution changesone nucleotide for another. D. Incorrect. Aframeshift mutation causes the entire codebeyond the mutation to shift. TAKS 2 Bio 6C

TAKS 2 Bio 4B, 6A, 6C

TAKS 2 Bio 6A


Texas SE page TK

The Genetics of Missing Teeth

Approximately one in five peo-ple are born without the

ability to develop a full set of teeth.One form of this condition, knownas hypodontia, is caused by anautosomal-dominant mutation.Therefore, each child of anaffected parent has a 50 percentchance of inheriting the condition.The genetic basis for autosomal-dominant hypodontia wasdiscovered by researchers atBaylor College of Medicine andThe University of Texas atHouston.

Identifying the Gene The researchers studied a Hou-ston family in which 21 membershad hypodontia. They comparedthe DNA of those 21 individuals tothat of 22 of their relatives who didnot have hypodontia. This com-

parison revealed a differencebetween the affected individu-als and their relatives in a smallregion of chromosome 14. In-cluded in that region is a genecalled Pax-9, which is requiredfor tooth formation in mice.When the researchers lookedspecifically at the Pax-9 genesequence in the Houston fam-ily, they found that all 21 familymembers who had hypodontiahad a mutation in the Pax-9 gene.Their unaffected relatives did nothave the mutation nor did 150 un-affected individuals outside thefamily.

Pinpointing the MutationA base-by-base analysis of themutated Pax-9 gene in this familyshowed that the gene containedan inserted cytosine nucleotide.

This insertion leads to a prematuretermination of translation and asmaller-than-normal protein. Thisis not the only mutation, however,that can result in hypodontia.When the same research groupstudied another family affected bythe condition, they found a differ-ent type of point mutation—a substitution—in the Pax-9 gene.This substitution leads to a non-functional protein.

Because the genetic message is read as a series of triplet nucleotides,insertions and deletions of one or two nucleotides can upset thetriplet groupings. Imagine deleting the letter C from the sentence“THE CAT ATE.” Keeping the triplet groupings, the message wouldread “THE ATA TE,” which is meaningless. A mutation that causesa gene to be read in the wrong three-nucleotide sequence is called aframeshift mutation.

Describe the effect a repressor has on the lacoperon when lactose is present. 6A 6B

Explain the role of transcription factors andenhancers in eukaryotic gene expression. 4B 6B

Differentiate between exons and introns.

Critical Thinking Evaluating SignificanceWhich type of mutation would have a greatereffect on the sequence of amino acids in a pro-tein, a base-pair substitution or a frameshiftmutation? Explain your answer. 4B 6A 6C

A mutation in which onenucleotide in a gene is replaced with a differentnucleotide is called 6C

A a deletion. C a substitution.B an insertion. D a frameshift mutation.


Section 2 Review

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Student EditionTAKS Obj 2 Bio 4B, 6A, 6CTAKS Obj 2 Bio 6B (grade 11)TEKS Bio 4A

Teacher EditionTAKS Obj 2 Bio 6A, 6B, 6CTEKS Bio 6A, 6B, 6C

pp. 220–221

AlternativeAssessmentHave students work in groups offour. Provide each group with a largeposter board. Ask each group tocreate its own DNA sequence bywriting the names of 24 bases on theposter. The first three bases must beTAG, and the last three bases mustbe either ATT, ATC, or ACT. Thebases in-between should be randombut when reading the bases astriplets, should not be any of thefour triplets indicated above. Haveeach group determine the mRNAsequence that would form and theprotein that would be produced.Next have the group choose a pointmutation—substitution, insertion, ordeletion—and “mutate” its DNA,and determine how the mutationaffects the protein. TAKS 2 Bio 6A

Co-op Learning

GENERAL


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

GENERAL

GENERAL


Answer to Concept Map

The following is one possible answer to Performance Zone item 15.

involves

of by on

inof a

producesproduces

begins at

organized as

organized as complementary to

correspond to

made of

made of

Gene expression

anticodons cytoplasmgene

translationtranscription

rRNA amino acidscodons

promoter mRNAtRNA ribosome proteins

Key Concepts

Study CHAPTER HIGHLIGHTS

ZONEKey Terms

Section 1ribonucleic acid (RNA) (208)uracil (208)transcription (208)translation (208)gene expression (208)RNA polymerase (209)messenger RNA (211)codon (211)genetic code (211)transfer RNA (212)anticodon (212)ribosomal RNA (212)

Section 2operator (216)operon (216)lac operon (216)repressor (216)intron (218)exon (218)point mutation (219)

BIOLOGYBIOLOGYUnit 6—Gene ExpressionUse Topics 3–6 in this unit to review the key concepts and terms in this chapter.

From Genes to Proteins

● The instructions needed to make proteins are coded inthe nucleotides that make up a gene. The instructionsare transferred to an mRNA molecule during transcrip-tion. The RNA is complementary to the gene, and theRNA nucleotides are put together with the help of RNApolymerase.

● During translation, the mRNA molecule binds to a ribo-some, and tRNAs carry amino acids to the ribosomeaccording to the codons on the mRNA. Each codon specifies an amino acid. The amino acids are joined toform a protein.

● The genetic code (codons) used by most organisms totranslate mRNA is nearly universal.

Gene Regulation and Structure

● Prokaryotic and eukaryotic cells are able to control whichgenes are expressed and which are not, depending on thecell’s needs.

● In prokaryotes, gene expression is regulated by operons.Gene expression is switched off when repressor proteinsblock RNA polymerase from transcribing a gene.

● In eukaryotes, an enhancer must be activated for a eukary-otic gene to be expressed. Transcription factors initiate tran-scription by binding to enhancers and to RNA polymerases.

● Many eukaryotic genes are interrupted by segments ofDNA that do not code for proteins; these segments arecalled introns. The segments of DNA that are expressed arecalled exons. After transcription, the introns are cut out,and the exons are joined. The exons are then translated.

● Mutations are changes in DNA. Gene alterations are muta-tions that change a gene. These mutations can involve achange in a single nucleotide or an entire gene.

2

1

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IPC BenchmarkReview

To prepare students for the TAKS, havestudents review Force and Motion:Speed and Momentum, Acceleration,and Work and Power TAKS Obj 5IPC 4A on pp. 1055–1056 of the IPCRefresher in the Texas AssessmentAppendix of this book.

ANSWERS

Using Key Terms1. c2. b3. a4. d5. a. A codon is a three-nucleotide

sequence of mRNA that codesfor a specific amino acid or astart or stop signal. An anti-codon is a three-nucleotidesequence of tRNA that iscomplementary to an mRNAcodon.

b. The protein-making instruc-tions coded in DNA aretranscribed into mRNA. ThetRNA molecules carry specificamino acids to the ribosomes;rRNA is a component ofribosomes, where proteins are made.

c. A promoter is a sequence ofDNA that signals the start oftranscription. In prokaryotes,the operator is a piece of DNAthat overlaps the promoter andacts as an on-off switch. Anoperon is a group of prokary-otic genes involved in the samefunction, together with theirpromoter and operator. Arepressor is a protein involvedin regulating prokaryotic geneexpression—it binds to anoperator and blocks RNApolymerase from binding to a promoter.

d. Exons are portions of aeukaryotic gene that are trans-lated into a protein. Introns arethe noncoding regions of aeukaryotic gene that areremoved before translation.

TAKS 2 Bio 6CTAKS 2 Bio 6CTAKS 2 Bio 4BTAKS 2 Bio 4B

Section Questions1 1, 2, 5, 6, 7, 8, 12, 15, 17, 20, 222 3, 4, 9, 10, 11, 13, 16, 18

Assignment Guide

222 Chapter10 • How Proteins Are Made

CHAPTER 10

Understanding Key Ideas

6. d7. b8. d9. c

10. a11. b12. RNA—DNA does not contain uracil.

13. Transposons can inactivate a gene or cause amutation. Exons do not affect genes—theyare the gene, or the parts of a gene that aretranslated. TAKS 2 Bio 6A; 6B (grade 11 only)

TAKS 2 Bio 6A

TAKS 2 Bio 6B (grade 11 only)TAKS 2 Bio 6B (grade 11 only)TAKS 2 Bio 6B (grade 11 only)TAKS 2 Bio 6ATAKS 2 Bio 6ATAKS 2 Bio 6A

14. The insertion mutation resulted in the pre-mature termination of translation. The substitution mutation led to the formation of a non-functional protein.

15. One possible answer to the concept map isfound on the bottom of the Study Zone page.TAKS 1 Bio/IPC 2C

TAKS 2 Bio 6C

Using Key Terms1. The making of RNA based on the sequence

of nucleotides in DNA is called 4Ba. DNA replication. c. transcription.b. translation. d. gene regulation.

2. The making of proteins from the informationcarried by mRNA is called 4B 9Ba. DNA replication. c. transcription.b. translation. d. gene regulation.

3. A change in the genetic code is called 6Ca. mutation. c. codon.b. operon. d. operator.

4. Mutations that change one nucleotide in agene are called a(n) 6Ca. operon mutation. b. codon mutation. c. repressor protein. d. point mutation.

5. For each set of terms, write one or moresentences summarizing informationlearned in this chapter.a. codon and anticodonb. mRNA, tRNA, and rRNAc. promoter, operator, operon, and repressord. exon and intron

6. Anticodons are found on ______ molecules. 6Aa. mRNA c. rRNA b. DNA d. tRNA

7. Unlike DNA, RNA contains 4A 6Aa. the sugar deoxyribose.b. the nitrogen base uracil.c. a phosphate group.d. nucleotides.

8. A short chain of DNA has the nucleotidesequence ATA CCG. Its complementarymRNA nucleotide sequence is 6Aa. TAT GCC. c. TUT GCC.b. UAU GCC. d. UAU GGC.

9. The lac operon allows a bacterium to buildthe proteins needed for lactose metabolismwhen 6Ba. glucose is present.b. lactose is absent. c. lactose is present.d. glucose is absent.

10. Transcription of lactose-metabolizing genesis blocked when the _____ is bound to theoperator. 6Ba. repressor c. inducerb. operon d. enhancer

11. In eukaryotes, gene expression can beregulated by 6Ba. mutations. b. transcription factors.c. repressors.d. operons.

12. Does the drawing below represent a strandof RNA or a strand of DNA? Explain youranswer. 6A

13. Compare the way trans-posons and exons affect genes. 6A 6B

14. What type of transla-tion error occurred in those individualswho possessed a mutated Pax-9 gene? 6C

15. Concept Mapping Make a conceptmap that shows the role of RNA in geneexpression. Try to include the followingwords in your map: transcription, transla-tion, mRNA, tRNA, rRNA, gene, promoter,codons, anticodons, proteins, amino acids,ribosome, and cytoplasm. 2C 3E

PerformanceZONE

CHAPTER REVIEW

U U G GC C CCC A U UAA

222

Review and AssessTAKS Obj 1 Bio/IPC 2CTAKS Obj 2 Bio 4B, 6A, 6B, 6CTEKS Bio 3D, 4B, 6A, 6B, 6CTEKS Bio/IPC 2C

pp. 222–223

Critical Thinking

16. When lactose enters the cell, thelac operon is activated and thenecessary enzymes needed tometabolize lactose are produced.If lactose is not present, then theenzymes are not produced, con-serving resources.

17. The results are probably correct.Different types of cells synthesizedifferent proteins, so they couldhave different mRNA molecules.

18. The classmate is correct. Exonsare the portions of a gene that aretranslated, while introns are not.Introns are cut out of the genebefore translation.

19. Chromosomal mutations involvesections of a chromosome (thou-sands of nucleotides). Point muta-tions involve changes in one or afew nucleotides. Chromosomalmutations are potentially moresevere, but both can lead to non-functional proteins that are poten-tially life threatening.

Alternative Assessments

20. Students’ reports will vary. Manyantibiotics inhibit bacterial pro-tein sysnthesis by combining withribosomal proteins. Erythromycinand chloramphenicol combinewith the 50S ribosomal subunit.The tetracyclines, streptomycin,gentamicin, kanamycin, and thenitrofurans combine with the 30Sribosomal subunit. Mupirocinand puromycin inhibit proteinsynthesis at the tRNA level.

21. Protein chemists use computermodels and genetic engineering todesign synthetic compounds. Theymay design drugs or other bioac-tive compounds. Protein chemistsattend college, followed by aresearch-based graduate program.Often they progress to a postdoc-toral appointment at a university.They are employed by researchorganizations, including universi-ties, and by private companies,such as drug manufacturers. Thegrowth potential of this field isgood. Starting salary will vary byregion.

22. See answer to question 20.TAKS 2 Bio/IPC 2D; TAKS 2 Bio 4B, 6B(grade 11 only)

Bio 3D

TAKS 2 Bio 4B

TAKS 2 Bio 6C

TAKS 2 Bio 6C


1. A. Incorrect. During transcription, mRNA ismade using DNA as a template. B. Correct.The parts represented in the drawing includemRNA, tRNA, ribosome, and growing proteinchain. C. Incorrect. Transformation is not aprocess associated with protein synthesis. D. Incorrect. DNA replication involves fewercomponents and looks like an open zipper.

2. F. Correct. The mRNA strand contains thecodons. G. Incorrect. B represents the anti-codon of the tRNA molecule. H. Incorrect. C represents the ribosome, site of protein


synthesis. J. Incorrect. D represents an aminoacid being brought to the growing proteinchain.

3. A. Incorrect. The ribosome is the site of pro-tein synthesis, shown as C. B. Correct. Each of the “balls” represents an amino acid thathas been added to the growing protein chain. C. Incorrect. The messenger RNA contains thecodons and is represented by A. D. Incorrect.The transfer RNA brings amino acids to thegrowing chain and is represented by B.TAKS 2 Bio 6B (grade 11 only)



Test

Critical Thinking16. Applying Information How does gene

regulation of the lac operon promote home-ostasis in intestinal E. coli bacteria?

17. Evaluating Results A molecular biologistisolates mRNA from the brain and from the liver of a mouse and finds that themRNA molecules are different from eachother. Can these results be correct or hasthe biologist made an error? Explain youranswer.

18. Evaluating an Argument A classmate states that damage to exons is very likely to affect the synthesis of a protein, whiledamage to introns is not. Evaluate thatstatement.

19. Evaluating Significance Compare andcontrast chromosomal mutations withpoint mutations, and evaluate the signifi-cance of each.

Alternative Assessment 20. Finding Information Use the media center

or Internet resources to learn about antibi-otics that interfere with protein synthesis.How do antibiotics fight bacterial infec-tion? Prepare an oral report that includesgraphics to interpret and summarize yourfindings.

21. Career Connection Protein ChemistResearch the field of protein chemistry, andwrite a report on your findings. Your reportshould include a job description, trainingrequired, kinds of employers, growthprospects, and a starting salary.

22. Interactive Tutor Unit 6 Gene ExpressionWrite a report summarizing how antibi-otics inhibit protein synthesis in bacteria.How do some antibiotics interfere withtranslation?


Use the model below and your knowledge ofscience to answer questions 1–3.

1. Which cellular function does this modelrepresent? A TranscriptionB TranslationC TransformationD DNA replication

2. Which part of the model represents acodon? F AG BH CJ D

3. What does the part labeled E represent? A RibosomeB Growing protein chainC Messenger RNAD Transfer RNA

Test questions are not necessarily arranged in orderof increasing difficulty. If you are unable to answer aquestion, mark it and move on to other questions.

E

A

C

B

D

3D

6B

6B

6B

6B

9A

6C

4B

2D 4B 6B

6C

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section 1 from genes to proteins - jourdanton · pdf file208 chapter 10 • how proteins...

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