Since its trailblazing First Edition, Biological Science has delivered numerous biology

teaching innovations that emphasize higher-order thinking skills and conceptual

understanding rather than an encyclopedic grasp of what is known about biology.

With each edition, this approach has grown and improved to better help students make the

shift from being novice learners to expert learners. Central to this shift is a student-centered

approach that provides deep support for the learning of core content and the development of

key skills that help students learn and practice biology.

Instruction Practice Application

Content

SkillsThinking likeT

a biologist

On the pages that follow, we will show how the text and MasteringBiology resources work

together to achieve this goal.

This model represents the

overarching goal of the Sixth

Edition: To help novice learners

progress from instruction . . .

. . . ultimately

completing the course

as expert learners who

think like biologists.

. . . and then to apply

what they have learned

to new situations . . .

. . . to become

active learners

through practice . . .

Chapter 17 Transcription, RNA Processing, and Translation 407

start translation anew. Termination occurs in very similar ways

in bacteria and eukaryotes.

Table 17.1 provided a comparison of transcription, RNA pro-

cessing, and translation as they occur in bacteria and eukaryotes.

What about the Archaea? It turns out that for transcription and

translation, archaea carry out these processes in ways much

more similar to eukaryotes than to bacteria.

Post-Translational Modifications

Proteins are not fully formed and functional when termination

of translation occurs. From earlier chapters, it should be clear

that most proteins go through an extensive series of processing

steps, collectively called post-translational modification, before

they are completely functional. These steps require a wide array

of molecules and events and take place in many different loca-

tions throughout the cell.

Folding A fundamental principle of biology is that a protein’s

function depends on its shape, and in turn, a protein’s shape

depends on how it folds (see Chapter 3). Folding is determined by

the amino acid sequence of a polypeptide chain. Although fold-

ing can occur spontaneously, it is frequently guided and acceler-

ated by proteins called molecular chaperones.

Chemical Modifications An earlier chapter described how

eukaryotic proteins are often extensively modified after they

are synthesized (see Chapter 7). For example, in the organelles

called the rough endoplasmic reticulum and the Golgi appa-

ratus, small chemical groups may be added to proteins—often

sugar or lipid groups that are critical for normal functioning.

Another common post-translational modification is the addition

of a phosphate group by enzymes called protein kinases. Adding

a phosphate group—and removing it later—often dramatically

affects the protein’s activity.

Figure 17.17 reviews how gene expression works in a eukary-

otic cell. Take a close look to see how all these steps work

together.

The take-home message is that gene expression is a multi-

step process that begins with transcription. What’s critical to

remember is that the RNAs and proteins produced during gene

expression give the cell and organism its characteristics. These

molecules truly are the stuff of life.

It turns out, however, that genes simply can’t be turned on all

the time. How does a cell “decide” which of its many genes should

be expressed and when to express them? These fundamental

questions are the focus of the next two chapters.

Nucleus

1. Transcription

2. RNA processing

Primary transcript(pre-mRNA)

TailCap

Mature mRNA

3. Translation

4. Post-translationalmodification (folding,glycosylation,transport, activation,degradation of protein)

mRNA

DNA

Ribosome

Pre-mRNA

Polymerase

Polypeptide

Active protein

Cytoplasm

Figure 17.17 The Major Steps of Gene Expression

in a Eukaryotic Cell.

CHECK YOUR UNDERSTANDING

If you understand that …

• Translation initiation occurs when (1) the ribosome binding

site on an mRNA binds to an rRNA sequence in the small

ribosomal subunit, (2) the initiator aminoacyl tRNA binds to

the start codon in the mRNA, and (3) the large subunit of

the ribosome attaches to the small subunit.

• Translation elongation occurs when (1) an appropriate

aminoacyl tRNA enters the A site, (2) a peptide bond forms

between the amino acid held by the tRNA in the A site and

the polypeptide held by the tRNA in the P site, and (3) the

ribosome moves down the mRNA one codon.

• Translation ends when the ribosome reaches a stop codon.

• Completed proteins fold, and in many cases are modified by

addition of chemical groups.

You should be able to …

1. explain why it’s important that the initiator trna be placed

in the p site instead of the a site.

2. explain why it’s logical that a release factor has the same

structure as an aminoacyl trna.

Answers are available in Appendix A.

408 Unit 3 Gene Structure and Expression

• In eukaryotes, transcription and translation of an RNA cannot

occur together because transcription occurs in the nucleus and

translation occurs in the cytoplasm.

• Transfer RNAs (tRNAs) serve as the chemical bridge between the

RNA message and the polypeptide product.

17.4 The Structure and Function

of Transfer RNA

• Each transfer RNA carries an amino acid corresponding to the

tRNA’s three-base-long anticodon.

• tRNAs have an L-shaped tertiary structure. One leg of the L con-

tains the anticodon, which forms complementary base pairs with

the mRNA codon. The other leg holds the amino acid specified by

that codon.

• Enzymes called aminoacyl-tRNA synthetases link the correct

amino acid to the correct tRNA.

• Because imprecise pairing—“wobble pairing”—can occur in the

third position of the codon and anticodon, the approximately 40

types of tRNA in the cell are enough to translate all 61 codons that

code for amino acids.

17.5 The Structure of Ribosomes

and Their Function in Translation

• Ribosomes are large macromolecular machines made of many

proteins and RNAs.

• In the ribosome, the tRNA anticodon binds to a three-base-long

mRNA codon to bring the correct amino acid into the ribosome.

• Peptide-bond formation by the ribosome is catalyzed by a ribo-

zyme (RNA), not an enzyme (protein).

• Protein synthesis occurs in three steps: (1) an incoming amino-

acyl tRNA occupies the A site; (2) the growing polypeptide chain is

transferred from a tRNA in the ribosome’s P site to the amino acid

bound to the tRNA in the A site, forming a peptide bond; and (3)

the ribosome moves to the next codon on the mRNA, accompanied

by ejection of the uncharged RNA from the E site.

• Chaperone proteins help fold newly synthesized proteins.

• Most proteins need to be chemically modified after translation

(post-translational modification) to activate them or target them

to specific locations.

17.1 An Overview of Transcription

• In transcription, RNA polymerase produces an RNA molecule with

a base sequence complementary to the base sequence of the DNA

template strand.

• RNA polymerase begins transcription by binding to promoter

sequences in DNA with the help of other proteins.

• In bacteria, this binding is accomplished through a protein called

sigma. Sigma associates with RNA polymerase and then recog-

nizes particular sequences within promoters that are centered 10

bases and 35 bases upstream from where transcription begins.

• Eukaryotic promoters vary more than bacterial promoters.

• In eukaryotes, transcription begins when a large array of proteins

called basal transcription factors bind to a promoter. In response,

RNA polymerase binds to the site.

• In bacteria and eukaryotes, RNA elongates in a 5′ S 3′ direction.

• Transcription in bacteria ends when a stem-loop structure forms

in the transcribed RNA; in eukaryotes, transcription terminates

after the RNA is cleaved downstream of the poly(A) signal.

17.2 RNA Processing in Eukaryotes

• In eukaryotes, the primary (initial) transcript must be processed

to produce a mature RNA.

• Splicing of primary transcripts removes stretches of RNA called

introns and joins together regions called exons.

• Complex macromolecular machines called spliceosomes splice

introns out of pre-mRNA.

• A “cap” is added to the 5′ end of pre-mRNAs, and a poly(A) tail is

added to their 3′ end.

• The cap and tail serve as recognition signals for translation and

protect the message from degradation by ribonucleases.

• RNA processing occurs in the nucleus.

17.3 An Introduction to Translation

• Ribosomes translate mRNAs into proteins with the help of adaptor

molecules called transfer RNAs.

• In bacteria, an RNA is often transcribed and translated at the same

time because there is no nucleus to separate these processes.

CHAPTER 17 REVIEW For media, go to MasteringBiology

Chapter 17 Transcription, RNA Processing, and Translation 409

Answers are available in Appendix A

TEST YOUR KNOWLEDGE

1. What does a bacterial RNA polymerase produce when it transcribes

a protein-coding gene?

a. rRNA

b. tRNA

c. mRNA

d. snRNA

2. CAUTION In bacteria, the initiation of translation starts at

a. the TATA box

b. the +1 site

c. the start codon, AUG

d. the Shine-Dalgarno sequence

3. Splicing begins:

a. as transcription occurs.

b. after transcription is complete.

c. as translation occurs.

d. after translation is complete.

4. Compared with mRNAs that have a cap and tail, predict what

will be observed when a eukaryotic mRNA lacks a cap and

poly(A) tail.

a. The primary transcript cannot be processed properly.

b. Translation occurs inefficiently.

c. Enzymes on the ribosome add back a cap and poly(A) tail.

d. tRNAs become resistant to degradation (being broken down).

TEST YOUR UNDERSTANDING

5. Which molecule would not be necessary for the expression of a

protein in an in vitro translation experiment performed by using

isolated components of the cell transcription machinery?

a. DNA polymerase

b. RNA polymerase

c. spliceosome

d. ribosomes

6. CAUTION A friend argues that redundancy of the genetic code

(see Chapter 16) is due to wobble pairing. Explain why this

isn’t the case.

7. Temperature-sensitive conditional mutations cause expression

of a wild-type phenotype at one growth temperature and a mutant

phenotype at another—typically higher—temperature. Imagine

that when a bacterial cell carrying such a mutation is shifted

from low to high growth temperatures, RNA polymerases in the

process of elongation complete transcription normally, but no new

transcripts can be started. The mutation in this strain most likely

affects what feature?

a. the terminator sequence

b. the start codon

c. sigma

d. one of the polypeptides of the core RNA polymerase

8. CAUTION In what ways are a promoter and a start codon similar?

In what ways are they different?

TEST YOUR PROBLEM-SOLVING SKILLS

9. The nucleotide shown below is called cordycepin triphosphate. It is a

natural product of a fungus that is used in traditional medicines.

O

OH3′

5′ BaseP P P

If cordycepin triphosphate is added to a cell-free transcription

reaction, the nucleotide is added onto the growing RNA chain but

no more nucleotides can then be added. The added cordycepin is

always found at the 3′ end of an RNA. Examine the structure of

cordycepin and explain why it ends transcription.

10. QUANTITATIVE A gene is composed of 5 exons: A, B, C, D, and E. How

many regions will not be translated in the pre-mRNA and the mRNA

of this gene? Calculate the cumulative size of the introns, keeping

in mind that the protein is made of 100 amino acids and the DNA

sequence from the start codon to the stop codon is 600 bp.

PUT IT ALL TOGETHER: Case Study

Amanita phalloides

What better not be for dinner?

Eating even a single death cap mushroom (Amanita phalloides) can

be fatal due to a compound called α-amanitin, a toxin that inhibits

transcription.

11. In the presence of α-amanitin, the level of mRNA in the cell

decreases. Which of the following enzymes is the least likely to be

inhibited by α-amanitin?

a. RNA polymerase

b. spliceosome

c. ribosome

d. enzyme-catalyzed addition of 5′ cap

410 Unit 3 Gene Structure and Expression

12. You have isolated a mutant of RNA polymerase II from a cancer cell

line with mutations in its active site. In an in vitro experiment, the

enzyme is functional and its activity is similar to the polymerase

found in non-cancerous cells. However, the mutant isolated from

the cancer is still sensitive to α-amanitin. Looking at the structure

presented in figure 17.3, propose a hypothesis to explain how the

α-amanitin could inhibit RNA polymerase II.

13. Toxins like α-amanitin are used for research in much the same

way as null mutants (see Chapter 16)—to disrupt a process

and see what happens when it no longer works. Researchers

examined the ability of α-amanitin to inhibit different RNA

polymerases. They purified RNA polymerases I, II, and III from

rat liver, incubated the enzymes with different concentrations

of α-amanitin, and then tested their activity. The results of

this experiment are shown below. These findings suggest that

α-amanitin-treated cells will have reduced levels of:

a. tRNAs

b. rRNAs

c. snRNAs

d. mRNAs

100

80

60

40

20

0.01 0.10 1.0 10

α-Amanitin concentration (μM)

Pe

rce

nt

po

lym

era

se

ac

tivit

y

Polymerase I

Polymerase III

Polymerase II

Source: Lindell, T. J., F. Weinberg, P. W. Morris, et al. (1970). Science 170: 447–449.

14. QUANTITATIVE If your aim was to use α-amanitin to shut down

95 percent of transcription by RNA polymerase II, roughly

what concentration of α-amanitin would you use? Note that the

scale on the x-axis of the graph in Question 13 is logarithmic

rather than linear, and each tick mark shows a tenfold

higher concentration. (See BioSkills 5 for tips on working with

logarithms.)

15. PROCESS OF SCIENCE Biologists have investigated how fast

pre-mRNA splicing occurs by treating cells with a toxin that

blocks the production of new pre-mRNAs, then following

the rate of splicing of the pre-mRNAs that were transcribed

before adding the toxin. Why was addition of a toxin

important in this study?

16. PROCESS OF SCIENCE The primary cause of death from

α-amanitin poisoning is liver failure. Suppose a physician

informs you that the liver cells die because their rate of

protein production falls below a level needed to maintain

active metabolism. Given that α-amanitin is an inhibitor

of transcription, you wonder if this information is correct.

Propose an experiment to determine whether the toxin also

has an effect on protein synthesis.

Students Go to MasteringBiology for assignments, the eText, and the

Study Area with animations, practice tests, and activities.

Professors Go to MasteringBiology for automatically graded tutorials and

questions that you can assign to your students, plus Instructor Resources.

411

Unit

3G

en

e S

tr

UC

tU

re

an

D e

xp

re

SS

iOn

This false-color

micrograph shows

projections from

human intestinal

cells (blue) and E. coli

bacteria (yellow).

In the intestine, the

nutrients available to

bacteria constantly

vary. This chapter

explores how changes

in gene expression

help bacteria respond

to environmental

changes.

Control of Gene Expression in Bacteria

Imagine waiting eagerly to hear the opening lines of a wonderfully melodic symphony played by a

renowned orchestra. The crowd applauds as the celebrated conductor comes onstage and then hushes

as he takes the podium. He cocks the baton; the musicians raise their instruments. As the baton comes

down, every instrument begins blaring a different tune at full volume. A tuba plays “Dixie,” a violinist

renders “Hey Jude,” a snare drum lays down beats for Daft Punk’s “Get Lucky,” while the bass drum simu-

lates cannons in the “1812 Overture.” Instead of music, there is pandemonium. The conductor staggers

offstage, clutching his heart.

Cacophony like this would result if a bacterial cell “played” all its genes at full volume all the time. The

Escherichia coli cells living in your gut right now have over 4300 genes. If all those genes were expressed

at the fastest possible rate at all times, the E. coli cells would stagger off the stage too. But this does not

happen. Cells are exquisitely selective about which genes are expressed, in what amounts, and when.

This chapter is part of the

Big Picture. See how on

pages 440–441.

18 In this chapter you will learn how

Bacteria turn their genes on and off

to adapt to changing environments

Different ways genes

can be regulated 18.1

How mutants help identify

regulated genes 18.2

Negative control of

gene expression

Positive control of

gene expression

Ways bacteria

regulate many genes

together

18.3

18.4

18.5

lookingcloserat

surveying

and