Chapter 17—From Gene to Protein...Chapter 17—From Gene to Protein “Let me out of the nucleus,...

Chapter 17—From Gene to

Protein

“Let me out of the nucleus, finally!”

or

The functional definition of a gene

I. The Connection Between Genes & Proteins

Metabolic defects = non-functional enzymes

• Studying metabolic diseases suggested that

genes specified proteins

- alkaptonuria (black urine from alkapton)

- PKU (phenylketonuria)

• Conclusion: Genes dictate phenotype

Beadle and Tatum (1941)

Experiments with Neurospora

ARGININE MUTANTS

One Gene-One Enzyme Hypothesis (Beadle & Tatum)

The function of a gene is to dictate the production of a specific enzyme

One Gene—One Enzyme� but not all proteins are enzymes� those proteins are coded by genes too

One Gene—One Protein�but many proteins are composed of several polypeptides, each of whichhas its own gene

One Gene—One Polypeptide

CENTRAL DOGMA of Molecular Biology

How does the information get from nucleus to cytoplasm?

Messenger RNA (mRNA)

= bridge between DNA & protein

DNA vs. RNA

RNA

DNA

StrandedNucleotidesSugar

Transcription—synthesis of RNA

under the direction of DNA• 1 DNA strand is template strand

• complementary RNA strand

is made– messenger RNA (mRNA)

• Enzyme = RNA polymerase

Translation:• Synthesis of a polypeptide under the

direction of mRNA

DNA → RNA → Protein

Prokaryotes vs. Eukaryotes

The Genetic Code

3 letter codon =

genetic “word”

Each word stands

for an amino acid

AUG = start codon

(starts a protein,

adds a methionine)

UAA/UAG/UGA =

stop codons

(signals end of

protein)

Code is universal—

from bacteria to

plants & animals (evolved very early

in the history of life)“Redundancy without Ambiguity”

II. The Synthesis & Processing of RNA

Transcription

Step 1 Initiation:RNA polymerase binds to promotersequence on DNA

Step 2 Elongation:RNA is made 5´→ 3´

Step 3 Termination:RNA polymerase stops at terminatorsequence

mRNA leaves nucleus through pores after processing

Transcription Animation

http://www-class.unl.edu/biochem/gp2/m_biology/animation/gene/gene_a2.html

Initiation of Transcription at a Eukaryotic Promoter

Transcription factors bind to

DNA

THEN…

RNA Polymerase binds

Transcription Factors =proteins which bind to DNA & turn transcription “on” or “off” (master regulators)

Transcription Unit = length of DNA that is transcribed into RNA

RNA Processing (only in eukaryotes!)

Step 1: Alteration of mRNA ends-- add 5’ cap (modified G (GTP))

-- protects RNA from degradation (hydrolytic enzymes in cytoplasm)-- “attach here” sign for ribosomes

-- add polyA tail (50-250 A’s)-- same as above ☺☺☺☺ and…-- helps mRNA export from nucleus

Step 2: RNA Splicing (cut & paste)-- Pre-mRNA → mRNA

-- edit out introns (noncoding regions)-- splice together exons (coding sequences)

In higher eukaryotes 90% or more of gene can be intronno one knows why…yet…

Spliceosome = snRNPs & proteins

� Recognize splice sites

� Ribozymes catalyze splicing process

Exon = protein domain

Alternative RNA splicing…

III. Synthesis of Protein

How are the codons read?

Translation Basics

� Ribosome reads mRNA in codons

� tRNA anticodon base pairs with codon of mRNA

� tRNA brings in correct amino acid

� Amino acids assembled into polypeptide chain

Structure of Transfer RNA (tRNA)

“clover leaf” structure

�anticodon on “clover leaf”end

�amino acid on 3’ end

Aminoacyl-tRNA Synthetases

Enzyme that bonds amino acids to tRNAs

Ribosomes

Consist of proteins and ribosomal RNA (rRNA)

2 subunits (large & small)

Made in nucleolus (eukaryotes) and then exported

Figure 17.15 pg. 316

P site (peptidyl-tRNA site)holds tRNA carrying growing polypeptide chain

A site (aminoacyl-tRNA site)holds tRNA carrying next amino acid to be added to chain

E site (exit site)discharged tRNA leaves ribosome from exit site

TRANSLATION DETAILSStep 1: Initiation—requires small ribosomal subunit, Met tRNA, large ribosomal subunit, initiation factors (proteins), & GTP (energy)

Step 2:

Elongation

Amino acids are added one by one to the preceding A.A.

Occurs in a 3-step cycle with the help of elongation factors (proteins):

1. Codonrecognition(A site)

2. Peptide bond formation

3. Translocation (A to P site)

Ribosome moves 5’ → 3’on mRNA

Step 3: Termination

stop codon in mRNA reaches A site

release factor (protein) binds to stop codon

protein is freed from the ribosome

other components disassemble

Polyribosomes—

Clusters of ribosomes

that translate a single mRNA simultaneously

Many copies of a protein are made

quickly

(prokaryotes &

eukaryotes)

Coupled Transcription & Translation

Prokaryotes—

With no nucleus bacteria can simultaneously transcribe & translate the same gene

Posttranslational Modifications—(Polypeptide → Functional Protein)

chemically modified amino acids, removal of amino acids, polypeptide cut into pieces, or several joined together

Signal peptide—(postal code)stretch of amino acids that targets the protein to a specific destination in a eukaryotic cell

Point mutations can affect protein

structure and function

Mutations—changes in the genetic material of a cell

Point mutation–a change in 1 base pair of a gene

Example—sickle cell anemia

2 Categories of point mutations:

1. Substitutions

silent mutations—

same A.A. inserted

missense mutations—

different A.A. inserted

nonsense mutations—

Stop codon formed

2. Insertions/Deletions

frameshift mutation—

changes reading frame

causes shortened,

nonfunctional proteins

Mutagens?

Copyright © 2005 Pearson Education, Inc. publishing as Benjamin Cummings

1. A mutation results in a defective enzyme a. In the following simple metabolic pathway, what would be

a consequence of that mutation?

a) an accumulation of A and no production of B and C

b) an accumulation of A and B and no production of C

c) an accumulation of B and no production of A and C

d) an accumulation of B and C and no production of A

e) an accumulation of C and no production of A and B

ca b

enzyme benzyme a


3. A portion of the genetic code is UUU = phenylalanine, GCC = alanine, AAA = lysine, and CCC = proline. Assume

the correct code places the amino acids phenylalanine,

alanine, and lysine in a protein (in that order). Which of the following DNA sequences would substitute proline for

alanine?

a) AAA-CGG-TTA

b) AAT-CGG-TTT

c) AAA-CCG-TTT

d) AAA-GGG-TTT

e) AAA-CCC-TTT


4.A particular triplet of bases in the coding sequence

of DNA is AAA. The anticodon on the tRNA that

binds the mRNA codon is

a) TTT.

b) UUA.

c) UUU.

d) AAA.

e) either UAA or TAA, depending on wobble in the first base.


5. A part of an mRNA molecule with the following sequence is

being read by a ribosome: 5' CCG-ACG 3' (mRNA). The

following activated transfer RNA molecules are available. Two of them can correctly match the mRNA so that a dipeptide can

form.

The dipeptide that will

form will be

a) cysteine-alanine.

b) proline-threonine.

c) glycine-cysteine.

d) alanine-alanine.

e) threonine-glycine.

CysteineACG

AlanineCGG

GlycineCCG

ThreonineUGC

AlanineCGU

ProlineGGC

Amino AcidtRNA Anticodon


7. Each of the following is a modification of the sentence THECATATETHERAT.

A. THERATATETHECATB. THETACATETHERATC. THECATARETHERATD. THECATATTHERATE. CATATETHERAT

Which of the above is analogous to a frameshift mutation?

a) A

b) B

c) C

d) D

e) E


8. Each of the following is a modification of the sentence THECATATETHERAT.

A. THERATATETHECATB. THETACATETHERATC. THECATARETHERATD. THECATATTHERATE. CATATETHERAT

Which of the above is analogous to a single substitution mutation?

a) A

b) B

c) C

d) D

e) E


9. What is the relationship among DNA, a gene, and a chromosome?

a) A chromosome contains hundreds of genes, which are

composed of protein.

b) A chromosome contains hundreds of genes, which are

composed of DNA.

c) A gene contains hundreds of chromosomes, which are

composed of protein.

d) A gene is composed of DNA, but there is no relationship

to a chromosome.

e) A gene contains hundreds of chromosomes, which are

composed of DNA.


10.A biologist inserts a gene from a human liver cell into

the chromosome of a bacterium. The bacterium then transcribes this gene into mRNA and translates the mRNA into protein. The

protein produced is useless.

The biologist extracts the protein and mature mRNA that codes

for it. When analyzed you would expect which of the following

results?

a) the protein and the mature mRNA are longer than in human

cells

b) the protein and mature mRNA are shorter than expected

c) the protein is longer and the mRNA is shorter than expected

d) the protein is shorter and the mRNA is longer than expected

Chapter 17—From Gene to Protein...Chapter 17—From Gene to Protein “Let me out of the nucleus,...

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