AP Gene Expression from Gene to Protein

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From Gene to Protein

Chapter 17

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The work of researchers in
1920’s to 1950’s led to the
discovery of DNA as the

macromolecule responsible for

storing and transmitting

heredity information

Fig. 16-7

(c) Space-filling model

Hydrogen bond
3′ end

5′ end

3.4 nm

0.34 nm

3′ end

5′ end

(b) Partial chemical structure

(a) Key features of DNA structure

1 nm

How does this molecule

give organisms’ their

characteristics?

Archibald Garrod (1857 –

1936)

💧

A physician who noticed that
a rare condition called
Alkaptonuria had the same
pattern of recessive
inheritance as described by
Gregor Mendel

💧

He was the first to suggest that
genes dictate phenotypes

💧

He believed that the
sufferers of this disorder lacked
an enzyme important in a
chemical pathway

THREE DECADES LATER,

MEET

GEORGE BEADLE

Geneticist

EDWARD TATUM

Biochemist

They just loved Drosophila,
better known as Fruit Flies

They believed that the

different eye colors were
due to mutations in genes

that produced enzymes
important for producing

pigments

Drosophila are complicated

compared to Neurospora

crassa

Bread Mold

SEM of Bread Mold

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Multiple Choice

enyme x enxyme y

I -----------------------> 2 --------------------->3 

The figure shows a simple metabolic pathway. According to Beadle and Tatum's hypothesis, how many genes are necessary for this pathway?

• 2

• 3

• 5

• this can't be determined from the information provided

Multiple Choice

enzyme x. enzyme y. enzyme z

1---------------->2------------------3------------------->4

Refer to the metabolic pathway illustrated. If I, II, III, and IV are all required for growth, a bacterial strain that does not make enzyme X would be able to grow on medium supplemented with which of the following nutrient(s)?

• nutrient A only

• nutrient B only

• nutrients A and B only

• nutrients B, C, and D only

Multiple Choice

enzyme x. enzyme y. enzyme z

1---------------->2------------------3------------------->4

Refer to the metabolic pathway illustrated. If I, II, III and IV are all required for growth, a bacterial strain that does not make enzyme Y would be able to grow on medium supplemented with which of the following compounds?

• nutrient A only

• nutrient C only

• nutrients A and C only

• nutrients B and D only

We have modified this hypothesis

from

One gene – One enzyme
One gene – One protein

One gene – One polypeptide!

Basic principals behind the
central dogma of molecular
biology lie in the “arrows”.

DNA --> RNA --> PROTEIN

TRANSCRIPTION

AND

TRANSLATION

Open Ended

Write down as many differences between DNA and RNA that you can remember

DIFFERENCES

DNA

💧

DOUBLE STRANDED

💧

SUGAR DEOXYRIBOSE

💧

HAS THYMINE AS A BASE

💧

REMAINS IN NUCLEUS

RNA

💧

SINGLE STRANDED

💧

SUGAR RIBOSE

💧

HAS URACIL AS A BASE

💧

EXITS THE NUCLEUS

Fig. 17-3
💧

Prokaryotes can
perform transcription
and translation
simultaneously
because there is no
nuclear envelope

TRANSCRIPTION

TRANSLATION

DNA

mRNA

Ribosome

Polypeptide

(a) Bacterial cell

Nuclear
envelope

TRANSCRIPTION

RNA PROCESSING
Pre-mRN
A

DNA

mRNA

TRANSLATION

Ribosome

Polypeptide

(b) Eukaryotic cell

Eukaryotes

Transcription

occurs

in the nucleus
Primary transcript

is

processed

Translation

occurs
in the
cytoplasm

The Conundrum of the

Genetic Code

💧

If the genetic code is carried on DNA

💧

And if there are only 4 different nucleotides found on
DNA

💧

And if there are 20 different Amino Acids that make up
proteins

💧

Then how does this work?

Multiple Choice

If each nucleotide coded for a different amino acid, then only _________ amino acids would be specified.

• 2

• 4

• 6

• 8

Multiple Choice

If the genetic code was determined by 2 nucleotide combinations (AT, TA, CA, AC. TG, GT….) then _________________ amino acids would be specified.

• 4

• 8

• 16

• 32

Multiple Choice

If the genetic code was determined by 3 nucleotide combinations (ATT, TAG, …) then ___________ amino acids would be specified.

• 8

• 16

• 32

• 64

1961 Marshall

Nirenberg
cracks the

genetic code

He synthesized artificial
mRNA and added this to a
test tube containing amino
acids, ribosomes and the
required enzymes.

His first find was that UUU
mRNA codon specified
phenylalanine

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Fig. 17-5

Second mRNA base

First mRNA base (5′ end of codon)

Third mRNA base (3′ end of codon)

Multiple Choice

Which of the following best describes the redundancy in the genetic code?

• More than one codon can specify the addition of the same amino acid.

• The genetic code is universal (the same for all organisms).

• The genetic code is different for different domains of organisms.

• A single codon can specify the addition of more than one amino acid.

Fig. 17-4
💧

THE TRIPLET CODE

DNA
molecule

Gene 1

Gene 2

Gene 3

template strand->

TRANSCRIPTION

TRANSLATION

mRNA

Protein

Codon

Amino acid

coding strand->

Multiple Choice

The genetic code is essentially the same for all organisms. From this, one can logically assume which of the following statements to be true?

• Different organisms have different types of amino acids.

• DNA was the first genetic material.

• A gene from an organism can theoretically be expressed by any other organism.

• The same codons in different organisms translate into different amino acids.

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Fig. 17-6

(a) Tobacco plant expressing
a firefly gene

(b) Pig expressing a
jellyfish gene

Universal Genetic Code

Multiple Choice

Which of the following processes occurs as part of transcription?

• DNA is replicated

• RNA is synthesized

• proteins are synthesized

• mRNA attaches to ribosomes

Transcription

DNA --> RNA

Let’s take a
closer look!

Three Basic Steps in

Transcription

Initiation

Elongation

Termination

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Fig. 17-7

Promoter

Transcription unit

RNA polymerase

binds to promoter

5′

5′

3′

3′

Initiation1

2

3

5′

5′

3′

3′

RNA
transcript

Template strand of DNA

Elongation

Rewound
DNA

5′

5′

5′

5′

5′

3′

3′
3′

3′

RNA
transcript
Termination

5′

5′

3′

3′

3′

5′

Completed RNA transcript

Template
strand of DNA

Direction of
transcription
(“downstream”)

3′ end

RNA nucleotides

Coding
strand of DNA

Elongation

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Fig. 17-8

A eukaryotic promoter
includes a TATA box

3′

1

2

3

Promoter

TATA box

Start point

Template

Template
DNA strand

5′

3′
5′

Transcription
factors

Several transcription factors must
bind to the DNA before RNA
polymerase II can do so.

5′

5′

3′

3′

Additional transcription factors bind to
the DNA along with RNA polymerase II,
forming the transcription initiation complex.

RNA polymerase II

Transcription factors

5′
5′

5′

3′

3′

RNA transcript

Transcription initiation complex

Multiple Choice

A particular triplet of bases in the template strand of DNA is 5′-AGT-3′. What would be the corresponding codon for the mRNA that is transcribed?

• 3′-UCA-5′

• 3′-ACU-5′

• 5′-AGT-3′

• 5′-UCA-3′

Fig. 17-9

💧

Facilitate the export of the mature mRNA

💧

Helps protect from hydrolytic degradation in
cytoplasm

💧

Helps ribosomes attach to the 5’ end

Protein-coding segment Polyadenylation signal

3′

3′ UTR

5′ UTR

5′

5′ Cap

Start codon

Stop codon

Poly-A tail

GP

PP

AAUAAA

AAA

AAA…

Cap and

Tail

Eukaryotes
tailor RNA
transcripts

RNA is modified after
transcription during RNA
processing

Both ends of the primary
transcript are altered

Certain sections of the
transcript are cut out
(introns) and the desired
sections (exons) are spliced
together

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Fig. 17-10

Pre-mRN
A

mRNA

Coding
segment

Introns cut out and
exons spliced together

5′ Cap

Exon Intro

n

5′

1

30

31

104

Exon

Intro
n

105

Exon

146

3′

Poly-A
tail

Poly-A
tail

5′ Cap

5′ UTR

3′ UTR

1

146

RNA

Splicing

Multiple Choice

Which of the following is most likely to occur if a researcher removes the 5′ cap and poly-A tail of an mRNA and inserts the mRNA molecule into a eukaryotic cell?

• The mRNA will be quickly converted into a ribosomal subunit.

• The cell will add a new poly-A tail to the mRNA.

• The mRNA will attaches to a ribosome and be translated, but more slowly.

• The molecule will be degraded by enzymes.

Transcription

Prokaryotes

💧

Transcription occurs in
cytoplasm

💧

One type of RNA polymerase

💧

RNA transcript is immediately
usable as mRNA

💧

RNA pol recognizes and binds to
the promoter

💧

Transcription ends at the
Terminator sequence

Eukaryotes

💧

Transcription occurs in nucleus

💧

Several types of RNA polymerase

💧

The RNA transcript undergoes
processing before it is sent out
of the nucleus

💧

Transcription factors must first
bind to the promoter, which
contains a TATA sequence, and
then RNA pol II binds

💧

Transcription ends at
polyadenylation signal sequence

Multiple Choice

Which of the following statements best describes the significance of the TATA box in the promoters of eukaryotes?

• It is the recognition site for the binding of a specific transcription factor.

• It sets the reading frame of the mRNA during translation.

• It is the recognition site for ribosomal binding during translation.

• It is the recognition site for ribosomal binding during transcription.

EUREKA! We
no longer think
that all enzymes

are proteins
Some RNA

functions just
like an enzyme

and we call

them ribozymes

RNA Splicing
by snurps
(snRNP’s)

•snRNP’s are small nuclear
ribonucleoproteins that
are located in the nucleus

•They are made up of
proteins and short
sequences of RNA called
snRNA (small nuclear
RNA)

•These snurps come
together with some other
proteins to form a
spliceosome which
“tailors” the pre-RNA

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Fig. 17-11-3

RNA transcript (pre-mRNA)

Exon 1

Exon 2

Intron

Protein
snRNA

snRNPs

Other

proteins

5′

5′

Spliceosome

Spliceosome
components

Cut-ou
t
intron
mRNA

Exon 1

Exon 2

5′

Multiple Choice

Which of the following statements correctly describes a ribozyme?

• It is an enzyme that catalyzes the association between the large and small ribosomal subunits.

• It is a catalyst that uses RNA as a substrate.

• It is an RNA with catalytic activity.

• It is an enzyme that synthesizes RNA as part of the transcription process.

Evolutionary Significance of

RNA Processing –

Alternative RNA Splicing

💧

DNA non-coding regions –
which codes for Introns might
be important for regulation of
how much protein to make,
when and where and
important structurally

💧

An organism can make many
different protein products
from one gene (alternative
RNA splicing)

💧

Exon shuffling could lead to
new proteins with new
possibly beneficial function

Multiple Choice

Which of the following processes correctly describes alternative RNA splicing?

• t is a mechanism for increasing the rate of translation.I

• It can allow the production of proteins of different sizes and functions from a single mRNA.

• It can allow the production of similar proteins from different RNAs.

• It increases the rate of transcription.

Multiple Choice

According to the central dogma, which of the following intermediate molecules correctly completes the flow chart describing the flow of information in a cell blank?

DNA →  ________ →  Proteins

• enzymes

• rRNA

• mRNA

• tRNA

Translation – the RNA –

directed synthesis of a

polypeptide

💧

rRNA – synthesized in the
region known as the
nucleolus, it combines with
special proteins to form
ribosomes and sent out to the
cytoplasm

💧

Cytoplasm stocked with all 20
amino acids and required
enzymes

💧

mRNA- carries the “recipe” for
making polypeptides out of
the nucleus. The genetic code
is determined by triplets of
nucleotides or codons

💧

tRNA – There are 20 different
types, each type carries one of
the specific amino acids to the
mRNA-ribosomal complex as
directed by the codons

Overview of
Translation

Examine this

diagram.

What can you

identify?

Fig. 17-14a

• tRNA

translates from
“language” of
nucleic acid to
amino acid

Amino acid
attachment site

(a) Two-dimensional structure

Hydrogen
bonds

Anticodon

3′

5′

Gives tRNA
it’s 3D shape

Binds to Codon
of mRNA

tRNA

Wobble

•There are 61 different
codons, yet only 45
different types of tRNA

•Wobble explains this:
Only the first two bases
are specific, the third not
so

•For instance, the tRNA
anticodon 3 to 5 reading
UCU can bind to the
codons reading 5 to 3 AGA
or AGG – but they both
are the codes for arganine

Fig. 17-14

Amino acid
attachment site

3′

5′

Hydrogen
bonds

Anticodo
n
(a) Two-dimensional structure

Amino acid
attachment site
5′

3′

Hydroge
n
bonds

3′

5′

Anticodo
n

Anticodo
n
(c) Symbol used
in this book

(b) Three-dimensional structure

ATP and

aminoacyl-tRNA
synthetase are

required to
attach amino

acid to tRNA

Charging
the tRNA

Multiple Choice

If a bacteria contained a mutant aminoacyl-tRNA synthetase that attaches a lysine instead of the normal phenylalanine to tRNAs with the anticodon AAA which of the following would occur?

•  None of the proteins in the cell will contain phenylalanine.

• Proteins in the cell will include lysine instead of phenylalanine at amino acid positions specified by the codon UUU.

• The cell will compensate for the defect by attaching phenylalanine to tRNAs with lysine-specifying anticodons.

• The ribosome will skip a codon every time a UUU is encountered.

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Fig. 17-16

Growing
polypeptide

Exit tunnel

Large
subunit

Small
subunit

tRNA
molecules

E P A

mRNA
5′

3′

(a) Computer model of functioning ribosome

P site (Peptidyl-tRNA
binding site)

E site
(Exit site)

A site (Aminoacyl-
tRNA binding site)

E P

A

Large
subunit

mRNA
binding site
Small
subunit

(b) Schematic model showing binding sites

Amino end

Growing polypeptide

Next amino acid
to be added to
polypeptide chain

mRNA
tRNA

E

3
′

5
′

Codons

(c) Schematic model with mRNA and tRNA

rRNA

• Made up of 2

subunits

one large, one small
• Eukaryotes have

much larger
ribosomes than
prokaryotes, and
they have slight
chemical differences
which are important
for medicine that
targets one without
affecting the other
(antibiotics)

• Huge ribozyme!

Three main stages of

Translation

💧

Initiation

💧

Elongation

💧

Termination

💧

Sound familiar?

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Fig. 17-17

3′

3′

5′

5′

U

U
A

A

C
G

Me
t

GTP

GDP
Initiato

r

tRNA
mRNA
5′
3′
Start codon

mRNA binding site

Small
ribosomal
subunit

5′

P site

Translation initiation complex

3′

E

A

Me
t

Large
ribosomal
subunit

INITIATION

Important for establishing the proper reading

frame

ELONGATION

CODON RECOGNITION

Charged tRNA base pairs to the A

site

PEPTIDE BOND FORMATION

rRNA molecule from large ribosome

catalyzed peptide bond. Growing
polypeptide attached to tRNA in A

site

TRANSLOCATION

tRNA in the A site moves to the P

site, the empty tRNA moves to E site

and is released

Fig. 17-18-4
💧

ELONGATION

Amino end
of polypeptide

mRNA

5′

3′
E

P

site
A
site

GTP

GDP

E

PA

E

PA

GDP

GTP

Ribosome ready for
next aminoacyl tRNA

E

PA

Fig. 17-19-3

Release
factor

3′

5′

Stop codon
(UAG, UAA, or UGA)

5′

3′
2

Free
polypeptid
e

2 GDP

GTP

5′

3′

Terminatio

n

The release factor is a protein shaped like tRNA, but doesn’t carry an amino

acid

It frees the completed polypeptide and the whole complex disassociates

Multiple Choice

In eukaryotes which of the following is the first step in translation?

• base pairing of activated methionine-tRNA to AUG of the messenger RNA

• binding of the larger ribosomal subunit to smaller ribosomal subunits

• the ribosome reaches a stop codon

• the small subunit of the ribosome recognizes and attaches to the 5′ cap of mRNA

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Fig. 17-25

TRANSCRIPTION

RNA PROCESSING

DNA

RNA
transcript

3′

5′
RNA
polymerase

Poly-A

Poly-A

RNA transcript
(pre-mRNA)

Intro
n

Exon

NUCLEUS

Aminoacyl-tRNA
synthetase

AMINO ACID
ACTIVATION

Amin
o
acid
tRNA

CYTOPLASM

Poly-A

Growing
polypeptide

3′

Activated
amino acid

mRNA

TRANSLATION

Cap

Ribosomal
subunits

Cap

5′

E

P

A

A

Anticodo
n

Ribosome

Codon

E