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4: Recombination Takes Place Through the Breakage, Alignment, and Repair of DNA Strands

4: Recombination Takes Place Through the Breakage, Alignment, and Repair of DNA Strands

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DNA Replication and Recombination

A

4 Branch migration takes place
as the two nucleotide strands
exchange positions, creating
the two duplex molecules.
A

A

5 This view of the structure
shows the ends of the two interconnected duplexes pulled
away from each other.

6 Rotation of the bottom
half of the structure…
B

B
B

a

b

Heteroduplex DNA

Branch point
b
b
a
a

Holliday intermediate

7 …produces
this structure.

A

B

strand plus one new strand from the other DNA molecule.
The point at which nucleotide strands pass from one DNA
molecule to the other is the Holliday junction (see Figure
9.16). The junction moves along the molecules in a process
called branch migration. The exchange of nucleotide
strands and branch migration produce a structure termed
the Holliday intermediate, which can be cleaved in one of
two ways. Cleavage may be in the horizontal plane, followed
by rejoining of the strands, producing noncrossover recombinants, in which the genes on either end of the molecules
are identical with those originally present (gene A with gene
B, and gene a with gene b). Cleavage in the vertical plane,
followed by rejoining, produces crossover recombinants, in
which the genes on either end of the molecules are different
from those originally present (gene A with gene b, and gene
a with gene B).

Horizontal
plane
Cleavage in the
horizontal plane…

✔ Concept Check 8
Why is recombination important?

Vertical
plane
a

Cleavage

Cleavage
…and rejoining
of the nucleotide
strands…

A

B

b

b

a

a

Noncrossover
(i) recombinants

Crossover
(k) recombinants

A

B

A

b

a

b

a

B

…produces noncrossover
recombinants consisting of
two heteroduplex molecules.

9.16 The Holliday model of homologous recombination. In
this model, recombination takes place through a single-strand break in
each DNA duplex, strand displacement, branch migration, and
resolution of a single Holliday junction.

…and rejoining
of the nucleotide
strands…

A

B

Concepts
Homologous recombination requires the formation of heteroduplex DNA consisting of one nucleotide strand from each of two
homologous chromosomes. In the Holliday model, homologous
recombination is accomplished through a single-strand break in
the DNA, strand displacement, and branch migration.

Cleavage in the
vertical plane…

b

…produces crossover
recombinants consisting of
two heteroduplex molecules.

Conclusion: The Holliday model predicts noncrossover
or crossover recombinant DNA, depending on whether
cleavage is in the horizontal or the vertical plane.

237

238

Chapter 9

Concepts Summary
• Replication is semiconservative: DNA’s two nucleotide strands




separate, and each serves as a template on which a new strand
is synthesized.
All DNA synthesis is in the 5¿ : 3¿ direction. Because the two
nucleotide strands of DNA are antiparallel, replication takes
place continuously on one strand (the leading strand) and
discontinuously on the other (the lagging strand).
Replication begins when an initiator protein binds to a
replication origin and unwinds a short stretch of DNA to
which DNA helicase attaches. DNA helicase unwinds the DNA
at the replication fork, single-strand-binding proteins bind to
single nucleotide strands to prevent secondary structures, and
DNA gyrase (a topoisomerase) removes the strain ahead of
the replication fork that is generated by unwinding.

• During replication, primase synthesizes short primers of RNA


nucleotides, providing a 3Ј-OH group to which DNA
polymerase can add DNA nucleotides.
DNA polymerase adds new nucleotides to the 3Ј end of a
growing polynucleotide strand. Bacteria have two DNA

polymerases that have primary roles in replication: DNA
polymerase III, which synthesizes new DNA on the leading
and lagging strands, and DNA polymerase I, which removes
and replaces primers.

• DNA ligase seals the nicks that remain in the sugar–phosphate





backbones when the RNA primers are replaced by DNA
nucleotides.
Several mechanisms ensure the high rate of accuracy in
replication, including precise nucleotide selection,
proofreading, and mismatch repair.
Precise replication at multiple origins in eukaryotes is ensured
by a licensing factor that must attach to an origin before
replication can begin.
The ends of linear eukaryotic DNA molecules are replicated by
the enzyme telomerase.
Homologous recombination takes place through breaks in
nucleotide strands, alignment of homologous DNA segments,
and rejoining of the strands.

Important Terms
semiconservative replication (p. 220)
equilibrium density gradient
centrifugation (p. 221)
replicon (p. 223)
replication origin (p. 223)
theta replication (p. 223)
replication bubble (p. 223)
replication fork (p. 223)
bidirectional replication (p. 223)
DNA polymerase (p. 225)
continuous replication (p. 225)
leading strand (p. 225)
discontinuous replication (p. 226)

lagging strand (p. 226)
Okazaki fragment (p. 226)
initiator protein (p. 226)
DNA helicase (p. 227)
single-strand-binding protein (SSB)
(p. 227)
DNA gyrase (p. 227)
primase (p. 228)
primer (p. 228)
DNA polymerase III (p. 229)
DNA polymerase I (p. 229)
DNA ligase (p. 230)

proofreading (p. 231)
mismatch repair (p. 232)
replication licensing factor (p. 232)
DNA polymerase ␣ (p. 233)
DNA polymerase ␦ (p. 233)
DNA polymerase ⑀ (p. 233)
DNA polymerase ␤ (p. 233)
DNA polymerase ␥ (p. 233)
telomerase (p. 234)
homologous recombination (p. 236)
heteroduplex DNA (p. 236)
Holliday junction (p. 236)

Answers to Concept Checks
1. Two bands
2. c
3. Initiator protein, helicase, single-strand-binding protein, DNA
gyrase
4. c
5. b

6. The size of eukaryotic genomes, the linear structure of
eukaryotic chromosomes, and the association of DNA with
histone proteins
7. c
8. Recombination is important for generating genetic
variation.

DNA Replication and Recombination

239

Worked Problems
1. The following diagram represents the template strands of a
replication bubble in a DNA molecule. Draw in the newly
synthesized strands and identify the leading and lagging strands.
Origin

2. Consider the experiment conducted by Meselson and Stahl in
which they used 14N and 15N in cultures of E. coli and equilibrium
density gradient centrifugation. Draw pictures to represent the
bands produced by bacterial DNA in the density-gradient tube
before the switch to medium containing 14N and after one, two,
and three rounds of replication after the switch to the medium
containing 14N. Use a separate set of drawings to show the bands
that would appear if replication were (a) semiconservative;
(b) conservative; (c) dispersive.

• Solution

• Solution
To determine the leading and lagging strands, first note which end
of each template strand is 5Ј and which end is 3Ј. With a pencil,
draw in the strands being synthesized on these templates, and
identify their 5Ј and 3Ј ends, recalling that the newly synthesized
strands must be antiparallel to the templates.

Origin













Unwinding


Unwinding

Origin
Next, determine the direction of replication for each new
strand, which must be 5¿ : 3¿ . You might draw arrows on the new
strands to indicate the direction of replication. After you have
established the direction of replication for each strand, look at each
fork and determine whether the direction of replication for a
strand is the same as the direction of unwinding. The strand on
which replication is in the same direction as unwinding is the
leading strand. The strand on which replication is in the direction
opposite that of unwinding is the lagging strand. Make sure that
you have one leading strand and one lagging strand for each fork.
Origin
Leading


Lagging




Lagging
Unwinding







Leading

Origin

Unwinding

DNA labeled with 15N will be denser than DNA labeled with 14N;
therefore 15N-labeled DNA will sink lower in the density-gradient
tube. Before the switch to medium containing 14N, all DNA in the
bacteria will contain 15N and will produce a single band in the
lower end of the tube.
a. With semiconservative replication, the two strands separate,
and each serves as a template on which a new strand is
synthesized. After one round of replication, the original template
strand of each molecule will contain 15N and the new strand of
each molecule will contain 14N; so a single band will appear in
the density gradient halfway between the positions expected of
DNA with 15N and of DNA with 14N. In the next round of
replication, the two strands again separate and serve as templates
for new strands. Each of the new strands contains only 14N, thus
some DNA molecules will contain one strand with the original
15
N and one strand with new 14N, whereas the other molecules
will contain two strands with 14N. This labeling will produce two
bands, one at the intermediate position and one at a higher
position in the tube. Additional rounds of replication should
produce increasing amounts of DNA that contains only 14N; so
the higher band will get darker.

Replication

Before the
switch
to 14N

Replication

After one
round of
replication

Replication

After two
rounds of
replication

After three
rounds of
replication

b. With conservative replication, the entire molecule serves as a
template. After one round of replication, some molecules will
consist entirely of 15N, and others will consist entirely of 14N; so
two bands should be present. Subsequent rounds of replication
will increase the fraction of DNA consisting entirely of new
14
N; thus the upper band will get darker. However, the
original DNA with 15N will remain, and so two bands
will be present.

240

Chapter 9

Replication

Replication

Replication

3. The E. coli chromosome contains 4.6 million base pairs of
DNA. If synthesis at each replication fork takes place at a rate of
1000 nucleotides per second, how long will it take to completely
replicate the E. coli chromosome with theta replication?

• Solution
Before the
switch
to 14N

After one
round of
replication

After two
rounds of
replication

After three
rounds of
replication

c. In dispersive replication, both nucleotide strands break down
into fragments that serve as templates for the synthesis of new
DNA. The fragments then reassemble into DNA molecules. After
one round of replication, all DNA should contain approximately
half 15N and half 14N, producing a single band that is halfway
between the positions expected of DNA labeled with 15N and of
DNA labeled with 14N. With further rounds of replication, the
proportion of 14N in each molecule increases; so a single hybrid
band remains, but its position in the density gradient will move
upward. The band is also expected to get darker as the total amount
of DNA increases.

Replication

Before the
switch
to 14N

Replication

After one
round of
replication

Replication

After two
rounds of
replication

Bacterial chromosomes contain a single origin of replication, and
theta replication usually employs two replication forks, which
proceed around the chromosome in opposite directions. Thus, the
overall rate of replication for the whole chromosome is 2000
nucleotides per second. With a total of 4.6 million base pairs of
DNA, the entire chromosome will be replicated in:
4,600,00 bp *

1 second
1 minute
= 2300 seconds *
2000 bp
60 seconds
= 38.33 minutes

At the beginning of this chapter, E. coli was said to be capable
of dividing every 20 minutes. How is this rate possible if it takes
almost twice as long to replicate its genome? The answer is that a
second round of replication begins before the first round has
finished. Thus, when an E. coli cell divides, the chromosomes that
are passed on to the daughter cells are already partly replicated. In
contrast, a eukaryotic cell replicates its entire genome once, and
only once, in each cell cycle.

After three
rounds of
replication

Comprehension Questions
Section 9.2
1. What is semiconservative replication?
*2. How did Meselson and Stahl demonstrate that replication in
E. coli takes place in a semiconservative manner?
*3. Draw a molecule of DNA undergoing replication. On your
drawing, identify (1) origin, (2) polarity (5Ј and 3Ј ends)
of all template strands and newly synthesized strands,
(3) leading and lagging strands, (4) Okazaki fragments,
and (5) location of primers.
4. Draw a molecule of DNA undergoing eukaryotic linear
replication. On your drawing, identify (1) origin,
(2) polarity (5Ј and 3Ј ends) of all template and newly
synthesized strands, (3) leading and lagging strands,
(4) Okazaki fragments, and (5) location of primers.
5. What are three major requirements of replication?
*6. What substrates are used in the DNA synthesis reaction?

Section 9.3
7. List the different proteins and enzymes taking part in
bacterial replication. Give the function of each in the
replication process.

8. What similarities and differences exist in the enzymatic
activities of DNA polymerases I, II, and III? What is the
function of each type of DNA polymerase in bacterial cells?
*9. Why is primase required for replication?
10. What three mechanisms ensure the accuracy of replication
in bacteria?
11. How does replication licensing ensure that DNA is
replicated only once at each origin per cell cycle?
*12. In what ways is eukaryotic replication similar to bacterial
replication, and in what ways is it different?
13. What is the end-of-chromosome problem for replication?
Why, in the absence of telomerase, do the ends of
chromosomes get progressively shorter each time the
DNA is replicated?
14. Outline in words and pictures how telomeres at the ends of
eukaryotic chromosomes are replicated.

DNA Replication and Recombination

241

Application Questions and Problems
Section 9.2
*15. Suppose a future scientist explores a distant planet and
discovers a novel form of double-stranded nucleic acid.
When this nucleic acid is exposed to DNA polymerases
from E. coli, replication takes place continuously on both
strands. What conclusion can you make about the structure
of this novel nucleic acid?
16. A line of mouse cells is grown for many generations in a
medium with 15N. Cells in G1 are then switched to a new
medium that contains 14N. Draw a pair of homologous
chromosomes from these cells at the following stages,
showing the two strands of DNA molecules found in the
chromosomes. Use different colors to represent strands with
14
N and 15N.
a. Cells in G1, before switching to medium with 14N
b. Cells in G2, after switching to medium with 14N
c. Cells in anaphase of mitosis, after switching to medium
with 14N
d. Cells in metaphase I of meiosis, after switching to
medium with 14N
e. Cells in anaphase II of meiosis, after switching to
medium with 14N
17. A bacterium synthesizes DNA at each replication fork at
a rate of 1000 nucleotides per second. If this bacterium
completely replicates its circular chromosome by theta
replication in 30 minutes, how many base pairs of DNA
will its chromosome contain?

Section 9.3
*18. The following diagram represents a DNA molecule
that is undergoing replication. Draw in the strands
of newly synthesized DNA and identify the following
items:
a. Polarity of newly synthesized strands
b. Leading and lagging strands
c. Okazaki fragments
d. RNA primers
Origin









Unwinding

Unwinding
Origin

*19. What would be the effect on DNA replication of mutations
that destroyed each of the following activities in DNA
polymerase I?
a. 3¿ : 5¿ exonuclease activity
b. 5¿ : 3¿ exonuclease activity
c. 5¿ : 3¿ polymerase activity
20. How would DNA replication be affected in a cell that is
lacking topoisomerase?
21. DNA polymerases are not able to prime replication, yet
primase and other RNA polymerases can. Some geneticists
have speculated that the inability of DNA polymerase to prime
replication is due to its proofreading function. This hypothesis
argues that proofreading is essential for faithful transmission
of genetic information and that, because DNA polymerases
have evolved the ability to proofread, they cannot prime DNA
synthesis. Explain why proofreading and priming functions in
the same enzyme might be incompatible?
22. A number of scientists who study ways to treat cancer have
become interested in telomerase. Why would they be
interested in telomerase? How might cancer-drug
therapies that target telomerase work?
23. The enzyme telomerase is part protein and part RNA. What
would be the most likely effect of a large deletion in the
gene that encodes the RNA part of telomerase? How would
the function of telomerase be affected?
24. Dyskeratosis congenita (DKC) is a rare genetic disorder
DATA characterized by abnormal fingernails and skin pigmentation,
the formation of white patches on the tongue and cheek, and
ANALYSIS
progressive failure of the bone marrow. An autosomal
dominant form of DKC results from mutations in the gene
that encodes the RNA component of telomerase. Tom
Vulliamy and his colleagues examined 15 families with
autosomal dominant DKC (T. Vulliamy et al. 2004. Nature
Genetics 36:447–449). They observed that the median age of
onset of DKC in parents was 37 years, whereas the median
age of onset in the children of affected parents was 14.5
years. Thus, DKC in these families arose at progressively
younger ages in successive generations, a phenomenon
known as anticipation. The researchers measured telomere
length of members of these families; the measurements are
given in the table on page 242. Telomere length normally
shortens with age, and so telomere length was adjusted for
age. Note that the age-adjusted telomere length of all
members of these families is negative, indicating that their
telomeres are shorter than normal. For age-adjusted
telomere length, the more negative the number, the shorter
the telomere.

242

Chapter 9

Age-Adjusted Telomere Length in Children and
Their Parents in Families with DKC
Parent telomere length
Ϫ4.7
Ϫ3.9
Ϫ1.4
Ϫ5.2
Ϫ2.2
Ϫ4.4
Ϫ4.3
Ϫ5.0
Ϫ5.3
Ϫ0.6
Ϫ1.3
Ϫ4.2

Child telomere length
Ϫ6.1
Ϫ6.6
Ϫ6.0
Ϫ0.6
Ϫ2.2
Ϫ5.4
Ϫ3.6
Ϫ2.0
Ϫ6.8
Ϫ3.8
Ϫ6.4
Ϫ2.5
Ϫ5.1
Ϫ3.9
Ϫ5.9

a. How do the telomere lengths of parents compare with
telomere length of their children?
b. Explain why the telomeres of people with DKC are
shorter than normal and why DKC arises at an earlier
age in subsequent generations.

Challenge Questions
Section 9.3
25. A conditional mutation expresses its mutant phenotype only
under certain conditions (the restrictive conditions) and
expresses the normal phenotype under other conditions (the
permissive conditions). One type of conditional mutation is
a temperature-sensitive mutation, which expresses the
mutant phenotype only at certain temperatures.
Strains of E. coli have been isolated that contain
temperature-sensitive mutations in the genes encoding
different components of the replication machinery. In each
of these strains, the protein produced by the mutated gene
is nonfunctional under the restrictive conditions. These
strains are grown under permissive conditions and then
abruptly switched to the restrictive condition. After one
round of replication under the restrictive condition, the
DNA from each strain is isolated and analyzed. What
characteristics would you expect to see in the DNA isolated
from each strain with a temperature-sensitive mutation in
its gene that encodes in the following?
a. DNA ligase
b. DNA polymerase I
c. DNA polymerase III

d. Primase
e. Initiator protein

26. DNA topoisomerases play important roles in DNA
DATA
replication and supercoiling (see Chapter 8). These
enzymes are also the targets for certain anticancer drugs.
ANALYSIS
Eric Nelson and his colleagues studied m-AMSA, one of the
anticancer compounds that acts on topisomerase enzymes
(E. M. Nelson, K. M. Tewey, and L. F. Liu. 1984. Proceedings
of the National Academy of Sciences 81:1361–1365). They
found that m-AMSA stabilizes an intermediate produced in
the course of the topoisomerase’s action. The intermediate
consisted of the topoisomerase bound to the broken ends
of the DNA. Breaks in DNA that are produced by anticancer
compounds such as m-AMSA inhibit the replication of the
cellular DNA and thus stop cancer cells from proliferating.
Propose a mechanism for how m-AMSA and other
anticancer agents that target topoisomerase enzymes
taking part in replication might lead to DNA breaks
and chromosome rearrangements.

10

From DNA to Proteins:
Transcription and RNA
Processing
RNA in the Primeval World

L

ife requires two basic functions. First, living organisms
must be able to store and faithfully transmit genetic
information during reproduction. Second, they must have
the ability to catalyze chemical transformations—to fire
the reactions that drive life processes. A long-held belief
was that the functions of information storage and chemical transformation are handled by two entirely different
types of molecules. Genetic information is stored in
nucleic acids. The catalysis of chemical transformations
was held to be the exclusive domain of certain proteins
that serve as biological catalysts or enzymes, making reactions take place rapidly within the cell. This biochemical
dichotomy—nucleic acid for information, proteins for
catalysts—revealed a dilemma in our understanding of the
early stages in the evolution of life. Which came first: proteins or nucleic acids? If nucleic acids carry the coding
instructions for proteins, how could proteins be generated
without them? Because nucleic acids are unable to copy
themselves, how could they be generated without proteins?
If DNA and proteins each require the other, how could life
begin?
This apparent paradox disappeared in 1981 when
Thomas Cech and his colleagues discovered that RNA can
serve as a biological catalyst. They found that RNA from the
protozoan Tetrahymena thermophila can excise 400
nucleotides from its RNA in the absence of any protein.
Molecular image of the hammerhead ribozyme (in blue) bound to RNA
(in orange). Ribozymes are catalytic RNA molecules that may have been the first
Other examples of catalytic RNAs have now been discovered
carriers of genetic information. [Kenneth Eward/Photo Researchers.]
in different types of cells. Called ribozymes, these catalytic
RNA molecules can cut out parts of their own sequences,
connect some RNA molecules together, replicate others, and
even catalyze the formation of peptide bonds between
amino acids. The discovery of ribozymes complements other evidence suggesting that the
original genetic material was RNA.
Ribozymes that were self-replicating probably first arose between 3.5 billion and 4 billion years ago and may have begun the evolution of life on Earth. Early life was an RNA
world, with RNA molecules serving both as carriers of genetic information and as catalysts
that drove the chemical reactions needed to sustain and perpetuate life. These catalytic
RNAs may have acquired the ability to synthesize protein-based enzymes, which are more

243

Chapter 10

efficient catalysts. With enzymes taking over more and more of the catalytic functions, RNA
probably became relegated to the role of information storage and transfer. DNA, with its
chemical stability and faithful replication, eventually replaced RNA as the primary carrier
of genetic information. In modern cells, RNA still plays a vital role in both DNA replication and protein synthesis.

(a)
5’
Strand
continues
–O

Phosphate
Base

O

P

O

C

O

N

4'
H

H

H

3’

2'
OH

O
–O

CH

Ribose
sugar

The Structure of RNA

N

H

C

H

N
H

H

H

OH

H

O

P

N

HC

O
H2C 5’

C

O
H

O

OH

H

H

N

C
HC

H

C

C

N

N

C

O

H

N
H

O

H
3’

H

C

H

O

H2C 5’

C

A

N

H

P

N
C

N

O

H
H

O

OH

Strand
continues
3’
(b) Primary structure
5’ AUGCGGCUACGUAACGAGCUUAGCGCGUAUACCGAAAGGGUAGAAC
An RNA molecule folds to
form secondary structures…

Folding

…owing to hydrogen bonding
between complementary
bases on the same strand.
3’

5’

CA

AG

A UG C
A UGCGGCUA CG

C

U

10.1 RNA has a primary and a secondary structure.

GAU

UC

Secondary
structure

A

AG

RNA, like DNA, is a polymer consisting of nucleotides joined
together by phosphodiester bonds (see Chapter 8 for a discussion of DNA structure). However, there are several
important differences in the structures of DNA and RNA.
Whereas DNA nucleotides contain deoxyribose sugars, RNA
nucleotides have ribose sugars (Figure 10.1a). With a free
hydroxyl group on the 2Ј-carbon atom of the ribose sugar,
RNA is degraded rapidly under alkaline conditions. The
deoxyribose sugar of DNA lacks this free hydroxyl group; so
DNA is a more stable molecule. Another important
difference is that thymine, one of the two pyrimidines found
in DNA, is replaced by uracil in RNA.
A final difference in the structures of DNA and RNA is
that RNA is usually single stranded, consisting of a single
polynucleotide strand (Figure 10.1b), whereas DNA normally consists of two polynucleotide strands joined by
hydrogen bonding between complementary bases. Although
RNA is usually single stranded, short complementary

H
H
3’
O

–O

G

N

O

H
3’

C

G

Before we begin our study of transcription, let’s review the
structure of RNA and consider the different types of RNA
molecules.

C

H2C 5’

–O

O
C

N

RNA has a hydroxyl group on
the 2’-carbon atom of its sugar
component, whereas DNA has
a hydrogen atom. RNA is more
reactive than DNA.

A

A

Strand of Ribonucleotides,
Participates in a Variety
of Cellular Functions

N

HC

O

10.1 RNA, Consisting of a Single

CH

1'
H

O

P

RNA contains uracil
in place of thymine.

C

U

O

H2C 5’

O

HN

G

he central dogma, first conceptualized by Francis Crick
(see Chapter 8), provides a molecular explanation for
how genotype encodes phenotype—that information in
DNA (genotype) is first transferred to RNA and then to protein (phenotype). This chapter is about the first steps in this
pathway of information transfer—the synthesis of an RNA
molecule through the process of transcription and the processing of that RNA. We begin with a brief review of RNA
structure and a discussion of the different classes of RNA. We
then consider the process of transcription, which synthesizes
an RNA molecule from a DNA template. Finally, we examine
the function and processing of RNA.

GA

T

AUGG
UACC

244

A C G

3'

From DNA to Proteins: Transcription and RNA Processing

Table 10.1

The structures of DNA and
RNA compared

Characteristic

DNA

RNA

Composed of
nucleotides

Yes

Yes

Type of sugar

Deoxyribose

Ribose

Presence of
2¿ -OH group

No

Yes

Bases

A, G, C, T

A, G, C, U

Nucleotides joined
by phosphodiester
bonds

Yes

Yes

Double or single
stranded

Usually double

Usually single

Secondary structure

Double helix

Many types

Stability

Stable

Easily degraded

regions within a nucleotide strand can pair and form secondary structures (see Figure 10.1b). Exceptions to the rule
that RNA is usually single stranded are found in a few RNA
viruses that have double-stranded RNA genomes.
Similarities and differences in DNA and RNA structures are
summarized in Table 10.1.

Classes of RNA
RNA molecules perform a variety of functions in the cell.
Ribosomal RNA (rRNA) along with ribosomal protein

Table 10.2

subunits make up the ribosome, the site of protein assembly. We’ll take a more detailed look at the ribosome later in
the chapter. Messenger RNA (mRNA) carries the coding
instructions for polypeptide chains from DNA to the ribosome. After attaching to a ribosome, an mRNA molecule
specifies the sequence of the amino acids in a polypeptide
chain and provides a template for joining amino acids. Large
precursor molecules, which are termed pre-messenger
RNAs (pre-mRNAs), are the immediate products of transcription in eukaryotic cells. Pre-mRNAs (also called primary transcripts) are modified extensively before
becoming mRNA and exiting the nucleus for translation
into protein. Bacterial cells do not possess pre-mRNA; in
these cells, transcription takes place concurrently with
translation.
Transfer RNA (tRNA) serves as the link between the
coding sequence of nucleotides in the mRNA and the amino
acid sequence of a polypeptide chain. Each tRNA attaches to
one particular type of amino acid and helps to incorporate
that amino acid into a polypeptide chain.
Additional classes of RNA molecules are found in the
nuclei of eukaryotic cells. Small nuclear RNAs (snRNAs)
combine with small protein subunits to form small nuclear
ribonucleoproteins (snRNPs, affectionately known as
“snurps”). Some snRNAs participate in the processing of
RNA, converting pre-mRNA into mRNA. Small nucleolar
RNAs (snoRNAs) take part in the processing of rRNA. A
class of very small and abundant RNA molecules, termed
microRNAs (miRNAs) and small interfering RNAs
(siRNAs), are found in eukaryotic cells and carry out
RNA interference (RNAi), a process in which these small
RNA molecules help trigger the degradation of mRNA or
inhibit their translation into protein. The different classes of
RNA molecules are summarized in Table 10.2.

Location and functions of different classes of RNA molecules

Class of RNA

Cell Type

Location of Function
in Eukaryotic Cells*

Ribosomal RNA (rRNA)

Bacterial and eukaryotic

Cytoplasm

Structural and functional
components of the ribosome

Messenger RNA (mRNA)

Bacterial and eukaryotic

Nucleus and cytoplasm

Carries genetic code for proteins

Transfer RNA (tRNA)

Bacterial and eukaryotic

Cytoplasm

Helps incorporate amino acids into
polypeptide chain

Small nuclear RNA (snRNA)

Eukaryotic

Nucleus

Processing of pre-mRNA

Small nucleolar RNA (snoRNA)

Eukaryotic

Nucleus

Processing and assembly of rRNA

MicroRNA (miRNA)

Eukaryotic

Cytoplasm

Inhibits translation of mRNA

Small interfering RNA (siRNA)

Eukaryotic

Cytoplasm

Triggers degradation of other RNA
molecules

*All eukaryotic RNAs are transcribed in the nucleus.

Function

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