Protein Synthesis
1. Background - How does DNA “control” a cell? Where are the instructions to make a protein? How does this happen?
a. Proteins carry information in their amino acid sequence while nucleic acids carry information in their nucleotide sequence.
b. Protein synthesis happens in two phases: (Fig. 16.3)
i. Transcription - the process of copying the instructions from the master DNA. Why is this necessary?
ii. Translation - converting the instructions from nucleic acid language to amino acid language.
iii. In eukaryotes, the product of transcription is modified before translation.
c. Problem: how can only 4 bases be used to indicate 20 amino acids?
i. Crick (yep, him again), 1961 - figured out how many bases have to be in each “code word” to create enough combinations to code for each amino acid; 41 = 4; 42 = 16; 43 = 64
ii. The code, therefore, is a series of triplets (three bases) which indicate a particular amino acid. The triplets are called codons. (Fig. 16.5)
iii. Note: an amino acid can be coded for by two or more codons but a given codon ALWAYS codes for only one amino acid. i.e. there is redundancy but no ambiguity. e.g., GAA and GAG both mean glutamic acid, but never any other amino acid. (Fig. 16.4)
iv. Evolutionary link - CCG codes for proline in every organism studied so far.
v. Remember, DNA is not permitted to leave the nucleus so a messenger is needed to take the information to the cytosol where the ribosomes are. That messenger is messenger RNA (mRNA).
2. Transcription (Fig16.7C) - the entire process is much like DNA replication except that DNA is acting as a template for the construction of mRNA rather than new for DNA. The DNA strand being transcribed is called the template strand. At some regions of DNA one strand is the template while at others the other strand is the template.
a. Initiation
i. RNA polymerase binds to DNA at regions called promoters, which include the initiation site (where transcription begins), and a few nucleotides upstream of the initiation site.
ii. In eukaryotes, the promoter is called a TATA box because it contains many A and T nucleotides. The TATA box is centered about 25 nucleotides upstream of the initiation site.
iii. Certain proteins, called transcription factors, must bind to the promoter region before RNA polymeras can bind (Fig. 16.8). Once RNA polymerase has bound, it separates the two strands about 10 bp at a time and begins transcribing.
b. Elongation
i. RNA polymerase moves along the template (coding) strand, catalysing the addition of bases according to base pairing rules to form mRNA
ii. Remember that in RNA, U rather than T is paired with A.
iii. The enzyme builds new RNA in the 5' ➝ 3' direction at a rate of about 60 nucleotides/s (sound familiar?).
iv. If there is high demand for a protein, the cell can have several RNA polymerases transcribing the same gene simultaneously to produce several mRNAs.
c. Termination
i. Transcription continues until RNA polymerase reaches the termination sequence (AATAAA).
ii. After some alterations, the mRNA is then sent to the cytosol.
iii. A transcription unit is the sequence between the start and stop sequences - generally speaking, one gene in eukaryotes. In prokaryotes, a transcription unit can be several genes that code for proteins of related functions.
3. Translation (Fig 16.9, 16.13, 16.14, 16.15).
a. Background
i. mRNA is translated into amino acid language by tRNA while the protein is built by a ribosome forming peptide bonds between amino acids.
ii. Each tRNA has an anticodon site which is a group of three nucleotides at one end of the tRNA. The anticodon binds by complementary base pairing to a specific codon of the mRNA.
iii. The tRNA carries the amino acid corresponding to the codon to which it’s anticodon is complementary. The amino acid attached to the tRNA at the amino acid attachment site.
b. Initiation
i. a ribosome has two tRNA binding sites called A and P. Imagine them as abbreviations for amino acid and polypeptide, respectively
ii. a ribosome binds to mRNA and begins looking for the start codon (AUG). As AUG also codes for methionine (Met), every polypeptide chain begins with that amino acid.
iii. when it is found, it is displayed in the P site so tRNA molecules can attempt to recognize it by complementary base pairing with their anticodon. When this occurs, the two ribosomal subunits come together to form the functional ribosome. The tRNA with the anticodon complementary to AUG always carries methionine (met) so it is always the first amino acid in every protein. The tRNA carrying methionine enters the P site
iv. the ribosome now displays the next codon in the A site and waits for the tRNA with the complementary anticodon to recognize it
v. Why can’t the ribosome begin translating anywhere? Codons are groups of three nucleotides but the ribosome can’t simply group in threes any old way. The nucleotides must be grouped correctly to form the right “words.” AUG serves to mark the starting point for reading the codons and, hence, determines how the codons are to be grouped. This establishes the reading frame - like spaces in our words.
vi. if the cellular demand for a protein is high, several ribosomes can translate the same mRNA simultaneously
c. Elongation (~ 60 ms per peptide bond)
i. the tRNA with an anticodon complementary to the next codon enters the A site
ii. the ribosome forms a new peptide bond by transferring the amino acid from tRNA in the P site to the amino acid on the tRNA in the A site
iii. tRNA (now empty) in the P site leaves and the ribosome moves over one codon on the mRNA
iv. the next codon is now displayed in the A site
v. the appropriate tRNA moves in and the ribosome attaches the amino acids (now 2 of them) from the tRNA in the P site to the amino acid on the tRNA in the A site. The chain is now three amino acids in length
vi. the polypeptide grows by the addition of one amino acid at a time in this manner
d. Termination
i. the elongation process continues until a stop codon is reached
ii. there is no tRNA which recognizes any of the stop codons
iii. the ribosome pauses briefly, waiting for a tRNA to enter the A site. It then severs the bond between the tRNA in the P site and the polypeptide
4. Mutations
a. Types of mutations
i. substitutions
(1) 1 nucleotide is replaced by another
(2) Are these mutations always harmful? Could give the same amino acid or one with similar properties; or the mutation could affect a non-critical part of the protein
(3) could be a problem if changed to a different amino acid or to a stop codon
ii. insertions or deletions
(1) addition or subtraction of one or more nucleotides
(2) usually serious because of frame shift unless three nucleotides are added or subtracted. Are these mutations always harmful? Could occur near the end of the mRNA, therefore affecting the end of the polypeptide
b. Mutagens
i. anything that increases the natural frequency of mutations
ii. X-rays, UV, other radiation can cause abnormal base pairing or irreversible bonding. Radiation can also break the ladder and splicing enzymes sometimes put the pieces back together in the wrong order
iii. often chemicals which have a shape similar to that of one of the nucleotides but makes different hydrogen bonds so that base pairing is not possible
iv. some chemicals change the shape of nucleotides so that base pairing is impossible or faulty
v. exposure to mutagens is especially dangerous for pregnant women because the embryo is growing rapidly and few embryonic cells give rise to all adult cells. A mutation in an embryonic cell will eventually affect many cells in the organism
vi. note the difference between mutations in somatic versus germ cell. Germ cells pass on mutations to all subsequent daughter cells. Somatic cell mutations are not hereditary
5. Gene therapy - useful to know where genes are so we can manipulate them
a. gene insertion
i. a normal copy of a gene is inserted into the correct position in DNA if a cell is missing it
ii. usually done by having a virus (called a vector) carry the gene into the cell
iii. e.g., nasal spray for cystic fibrosis
iv. e.g., insert gene for insulin into pancreatic cells and implant in patient. As new cells divide, pancreas wold slowly become non-diabetic
v. one problem is that a gene are often controlled by its location on the chromosome so we could over-produce a protein
vi. one problem is that genes are often controlled by their location on the chromosome so moving them may result in genes being out of control
b. gene transplant - defective gene can be removed and replaced by a functional one