Topic 13, Translation and Proteins
In translation, the language of mRNA is translated into the language of proteins
- Ribosome structure
A. Ribosomes are small bodies that contain about 90% of the RNA in the cell
(one bacterial cell contains ~10,000 ribosomes)
B. Ribosomes are the sites of protein synthesis in the cell
C. Ribosomes have two subunits, a large subunit and a small subunit
1. In prokaryotes, there are 50S and 30S subunits
a. 50S subunit has 23S rRNA, 5S rRNAs, and 31
proteins
b. 30S subunit has 16S rRNA and 21 proteins
c. The ribosomal subunits are capable of
self-assembly
2. In eukaryotes, there are 60S and 40S subunits
a. 60S subunit has 28S, 5.8S, and 5S rRNAs and
49 proteins
b. 40S subunit has 18S rRNA and 33 proteins
3. The ribosomes are ~1/2 protein and ~1/2 RNA
4. A single precursor molecule encodes all 3 rRNA sequences in E.
coli and there are
7 copies of this sequence in the genome
5. A single precursor molecule encodes the 18S, 28S, and 5.8S rRNAs in
eukaryotes,
the 5S is encoded elsewhere in the genome
a. All of the precursor molecules that encode
the 18S, 28S, and 5.8S rRNAs are at the
nucleolar organizing
region
b. There are hundreds to thousands of these
precursor molecule templates per haploid genome
c. The nucleolar organizing region is necessary
to organize the nucleolus,
no nucleolus is present
if the NOR is absent
d. The ribosomal subunits are made in the
nucleolus in eukaryotic cells
- tRNA structure
A. tRNAs are 75-90 nucleotides long, 4S
B. Transcribed as larger precursors and cleaved into mature 4S molecules
C. Holley et al. (1965) determined the complete nucleotide sequence of the first tRNA
D. The two-dimensional model has
1. An amino acid binding site (CCA sequence which binds the amino
acid)
2. Anticodon loop
3. D-loop and T-pseudouracil loop and a variable loop
4. It has several regions of paired bases and has the shape of a
cloverleaf
5. It contains many unusual bases
E. A three-dimensional model, given in Fig. 14.4
- Charging the tRNA (attaching the amino acid to the tRNA
A. Aminoacyl-tRNA synthetases or simply synthetases attach the correct amino acid
to the
correct tRNA in a two-step process (see handout for details)
1. Amino acid activation
2. Charging the tRNA with an amino acid --> aminoacyl-tRNA
- Translation
A. Initiation
1. Initiation complex - 3 components join together to form the
initiation complex
a. Small ribosomal subunit
b. Chain initiation codon in mRNA (AUG)
c. First tRNA attaches to the P (pepidyl-tRNA
binding) site on the small ribosomal subunit
d. The first AUG codon codes for formly
methionine in prokaryotes and methionine in eukaryotes.
Subsequent AUG codons
code for methionine in both prokaryotes and eukaryotes.
B. Elongation
1. The large ribosomal subunit joins the initiation complex
2. There are 3 major steps in elongation
a. A second aminoacyl-tRNA binds to the next
codon at the A site
b. A peptide bond forms between the amino acids
c. Translocation
1) The ribosome moves
one codon in relation to the mRNA
2) The tRNA moves from
the A site to the P site on the ribosome
3) The tRNA is released
from the P site on the ribosome
3. A tunnel exists within the large subunit through which the
elongating polypeptide emerges
C. Termination
1. The chain termination codons are UAA, UAG, and UGA
2. When a chain termination codon is reached, the large and small
subunits of the ribosome dissociate
3. The polypeptide chain and the mRNA are released from the ribosome
D. Therefore, the sequence of amino acids in a protein is determined by the sequence of
nucleotides
in the DNA
- Polyribosomes (=polysomes)
A. Several ribosomes are attached to each mRNA which simultaneously transcribe the mRNA
molecule
- Differences in translation in prokaryotes and eukaryotes
A. Eukaryotic mRNA is much longer-lived (hours) than prokaryotic mRNA (minutes)
B. The 5' end in eukaryotes is "capped" with a 7-methylguanosine residue and
a poly-U tail is added to the 3' end
C. The first amino acid is methionine in eukaryotes and formylmethionine in prokaryotes
- Garrod's initial evidence that genes are involved in protein production
A.Garrod (1902, 1908), an English physician, was interested in human diseases that might
have
a genetic basis
B.He studied the inheritance of alkaptonuria and concluded that it was due to an
autosomal recessive mutation
C. Characteristics of alkaptonuria
1. The affected individual's urine turns black when exposed to air
2. The cartridge in the ears and nose turns black and the whites of the
eyes turn black
3. These are all due to the accumulation of alkapton (homogentisic
acid)
D. Garrod fed affected individuals chemicals which are related to alkapton (phenylalanine
or tyrosine)
and found that the amount of alkapton in the urine increased.
However, when he fed these same
compounds to normal individuals, there was no increase of alkapton in
the urine.
E. Garrod concluded that there was a genetic block in the metabolism of homogentisic acid.
F. Garrod hypothesized that hereditary information controls chemical reactions in the body
G. Perhaps best stated, one gene-one metabolic block
- Beadle and Tatum's work provided the first convincing evidence that genes are directly
responsible for the production of proteins
A. Beadle and Tatum produced the first biochemical mutants (in Neurospora crassa, the
bread mold)
1. Neurospora will grow on a defined minimal medium, Fries
medium + biotin
a. Minimal medium only contains the nutrients
that are necessary for growth of a wild-type organism
1) If you remove any
one component of the minimal medium, the organism cannot grow or
grows
poorly
b. Complete medium is minimal medium
supplemented with other nutrients
2. Beadle and Tatum X-rayed conidia (asexual spores) to increase the
mutation frequency and
plated them on complete medium.
a. They then transferred cells of each colony
which grew onto minimal medium and
determined which ones
would not grow.
1) The ones which would
not grow were auxotrophs (nutritional-requiring mutants)
2) Ones that could grow
on minimal medium are referred to as prototrophs
b. They then placed the prototrophs
individually on minimal medium which was supplemented with
various classes of
compounds (e.g., amino acids, vitamins, purines and pyrimidines, etc.) to
determine which class
of compounds was necessary for growth of each mutant.
c. Once they determined the class of compound
which was affected in each mutant, they added the
individual members of
that class of compound one at a time to determine the specific one that
was necessary for
growth.
d. For example, if a auxotroph was not able to
grow in minimal medium but able to grow in
medium that was
supplemented with amino acids, they concluded that the organism was
not able to produce at
least one of the amino acids.
e. They then added the amino acids to minimal
medium one at a time to determine which amino
acid the organism could
not produce
2. They isolated hundreds of biochemical mutations and found that each
mutation had
a. A single nutritional requirement and
b. A mutation in only one gene
3. These observations led Beadle and Tatum to conclude that one gene
specifies one enzyme,
the one gene : one enzyme hypothesis
4. Beadle and Tatum were awarded the Nobel prize for this work in 1958
- Use of auxotrophs for working out biochemical pathways (Srb and Horowitz, 1944)
A. They isolated 7 auxotrophs in Neurospora crassa which each required arginine
for growth
B. They added arginine or presumptive precursors of arginine (ornithine or citrulline) to
minimal medium
and placed each auxotroph on the medium
1. One mutant grew if arginine was added but not if ornithine or
citrulline was added (arg-1 mutation)
2. Two mutants grew if arginine or citrulline were added but not if
ornithine was added
(arg-2 and 3 mutations)
3. Four mutants grew if arginine, citrulline, or ornithine were added
(arg 4, 5, 6, and 7 mutations)
C. From these observations, they concluded that a precursor was converted to ornithine
which was
converted to citrulline which was converted to arginine (see
notes and/or book for details)
D. The mutants within each group mapped at a single locus and each group of mutants mapped
to
different loci
E. In lab, we are working with different yeast mutations requiring adenine
1. Two adenine-requiring mutations were crossed together and plated on
minimal medium.
a. If the diploid progeny cells grew on minimal
medium, we would conclude that the two mutations
are not allelic (do
complement)
b. If the diploid progeny cells do not grow on
minimal medium, we would conclude that the two
mutations are allelic
(do not complement)
- A few examples of biochemical mutants in humans
A. Phenylketonuria = PKU
1. Due to a deficiency in the enzyme that converts phenylalanine to
tyrosine (occurs in the liver)
2. Autosomal recessive, occurs in 1/1100 live births
3. Causes severe mental retardation if not treated
4. The symptoms are due to the accumulation of phenylalanine
5. The developing babies with PKU are normal prior to birth because
excess phenylalanine is
removed by the placenta
6. Within 3 days after birth, the level of phenylalanine goes up
sharply in babies with PKU
7. If individuals with PKU are placed on a diet that is low in
phenylalanine for the first 6-9 years,
their development is relatively normal.
After this, they can have a normal diet because the brain is
fully developed by this time.
8. All newborns in the US are tested for PKU
B. Cretinism (due to an autosomal recessive)
1. An enzyme which is needed for synthesis of thyroxin by the thyroid
gland is defective in individuals
with this genetic disease
2. Fetal development is normal because the hormone comes from the
mother
3. A few weeks after birth, there is mental dullness and retarded
growth
4. If thyroxin (thyroid hormone) is administered, development is normal
C. Albinism
1. Due to an autosomal recessive
2. An enzyme is not produced which is necessary for the synthesis of
melanin
D. Tyrosinosis
1. Due to an autosomal recessive
2. Harmless condition in which the individual has an excess level of
tyrosine in the blood
E. Alkaptonuria, already discussed
F. Therefore, there are a large number of molecular diseases in this small biosynthetic
pathway
G. Galactosemia
1. Due to an autosomal recessive
2. 1/57,000 births
3. Enlarged liver, vomiting, cataracts, mental retardation, failure to
thrive (sickly)
4. Due to the lack of a liver enzyme, GPT (galactose 1-phosphate uridyl
transferase) which converts
galactose-1P to glucose-1P
5. The accumulation of galactose-1P causes the problem
6. If the infant is fed milk, the major carbohydrate in milk is the
dissacharide, lactose, which is a
glucose-galactose disaccharide
7. If detected at birth, the individual can be fed a diet that lacks
lactose to overcome this genetic
defect.
H. Enzymatic pathways
1. A precursor, A is converted by the action of an enzyme (enzyme A) to
endproduct B
2. If the enzyme is defective or missing, there will be an increased
level of compound A and a
decreased level of endproduct B.
3. For some genetic conditions, the accumulation of the precursor
causes the problem
a. Examples: galactosemia, PKU, alkaptonuria
4. For other genetic conditions, the lack of an endproduct causes the
problem
a. Examples: cretinism and albinism
5. Thousands of human genetic diseases are now known, and most are due
to an alteration in a single
gene. Many of these can be treated.
- Studies of human hemoglobin established that one gene encodes one polypeptide
A. Sickle cell anemia characteristics
1. Discovered in 1910 and inherited as a Mendelian recessive
2. Red blood cells (erythrocytes) in affected individuals assume a
distinctive sickle shape under
low oxygen conditions
3. These sickle cells plug capillaries resulting in oxygen loss to
tissues and the individual
experiences sickle-cell crisis. Can be
fatal.
4. Controlled by a single locus, HbA is the normal
allele and HbS is the mutant allele
B. Linus Pauling (1949) found that hemoglobin molecules from normal and affected
individuals had
different migration rates when subjected to electrophoresis. He
also found that the heterozygote
had both forms; thus, this gene exhibits codominance
C. Hemoglobin molecules have 4 iron-containing heme groups (non-protein) and 4 polypeptide
chains
(2 alpha globin chains and 2 beta globin chains)
D. Ingram (1954-1957)
1. Cut the alpha globin chains and beta globin chains into small pieces
(peptide fragments) and
2. separated the peptide fragments from each other by 2-dimensional
electrophoresis
3. He found that one peptide fragment from the beta globin chain was
different in normal and
sickle-cell individuals
4. Also found that there was a single amino acid substitution in this
peptide fragment
a. Normal had glutamic acid and sickle cell had
valine
5. This was the first demonstration that a mutation that affects a
phenotype can be due to a
single amino acid substitution
- Colinearity of a gene and a polypeptide (Yanofsky 1967)
A. He recovered many different mutations in the tryptophan synthetase gene in E. coli
B. He determined where each of the mutations caused an amino acid change in the
tryptophan synthetase protein
C. He also mapped each of the mutations within this gene
D. He found that the order of alterations in the polypeptide was the same as the
order of mutations in the genetic map: therefore,
the a gene and a polypeptide are collinear
- Protein chemistry.
I expect that each of you have had this in other courses and I expect you to know
important basic information about proteins such as:
A. There are 20 types of amino acids in living things
B. The amino acids are attached to each other by peptide bonds (which are covalent bonds)
C. A chain of amino acids is a polypeptide
D. There is a N-terminus and a C-terminus and you know what they are
E. There are 4 levels of protein structure that are recognized,
1. Primary structure = the sequence of amino acids in a polypeptide
2. Secondary structure = alpha helix or beta-pleated sheet
3. Tertiary structure = folding of a protein
4. Quaternary structure = association of polypeptides with each other
F. Proteins can be structural or enzymes
1. Enzymes = a protein or complex of proteins that catalyzes a specific
chemical reaction
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Updated 11/28/00