Week 7 proteins structure and synthesis

Question 1

Protein functions

Answer 1

> 50% of the cells mass
-Enzymatic
-Transport
-Defensive roles
-Regulation
-Structural support
-Receptor molecules
-Hormones
-Cell signalling
-Amino acid storage
-Movements within cells

Question 2

Protein features

Answer 2

Linear chains of amino acids, arranged in a 3D structural hierarchy

Question 3

Protein features 2

Answer 3

Diversity of structures, resulting in a wide range of
a functions:
▪ functions depend on the 3D protein shape
▪ 3D protein shape is determined by its
amino acid sequence
Without its normal conformation (native folding), the protein loses its biological activity

Question 4

Amino acids

Answer 4

Monomers
Proteins are assembled from a set of 20 different amino acids

Question 5

Amino acid structure

Answer 5

Amino acids have in common: Carbon α (Cα), amino
(-NH2) and carboxyl (–COOH) functional groups
R group or side chain is specific for each amino acid

Question 6

Polypeptides

Answer 6

Protein chains
In a protein, amino acids are linked together by covalent peptide bonds formed between carboxylic acid and amino groups of two consecutive amino acids

Question 7

Polypeptides 2

Answer 7

-Form linear, unbranched polypeptides (backbone of the protein)
-Every protein has a unique sequence of amino acids with a carboxyl end (C-term) and an amino end (N-term)

Question 8

4 levels of protein structure

Answer 8

The architecture of proteins has four levels of organisation:
1) Primary - amino acids
2) Secondary - α-helix / β-pleated sheet
3) Tertiary
4) Quaternary

Question 9

Primary structure

Answer 9

The unique sequence of amino acids in a polypeptide chain,
▪ Sequence is determined by the DNA of the gene that encodes the protein
▪ Linked by peptide bonds
-Peptide = short chain (<50 amino acids)
-Protein = longer chain (>50 amino acids).
Also referred to as polypeptides

Question 10

Secondary structure

Answer 10

Local and folded structures that form within a polypeptide due to:
- H- bonds between backbone of amino and carbonyl groups of two amino acids
Forms 2 structural patterns:
α-helix and β-pleated sheet

Question 11

Tertiary structure

Answer 11

Tertiary structure is the three-dimensional folding of a polypeptide
▪ determined by interactions among various side chains

Question 11

Quaternary structure

Answer 11

Quaternary structure is the result of the association of two or more polypeptide
chain subunits into a closely packed arrangement

Question 12

Protein Synthesis in eukaryotic mammalian cells

Answer 12

Protein synthesis (or gene expression) consists in two phases - transcription
and translation (separated by the RNA maturation in eukaryotic cells)
Transcription - in the nucleus
Translation - By ribosomes in the cytoplasm

Question 13

DNA transcription - key points

Answer 13

DNA to mRNA
-In eukaryotic cells, it occurs in the nucleus
-RNA polymerase uses one of the DNA strands as a template to produce pre-messenger RNA (pre-mRNA) with a complementary sequence to DNA.
-It adds RNA nucleotides (complementary to DNA) in the 5′→3′ direction
Consists of 3 phases:
-Initiation
-Elongation
-Termination

Question 14

Transcription - initiation

Answer 14

-RNA polymerase, with the help of transcription factors, binds to a specific sequence of DNA (promoter – just before the start site of a gene)
-RNA polymerase separates the double strands and use a single strand as a template
-Begins the complementary RNA strand synthesis

Question 15

Transcription - elongation

Answer 15

-RNA polymerase moves along the DNA coding sequence in 5ʹ-to-3ʹ direction.
-Melting DNA and adding RNA nucleotides

Question 16

Transcription - termination

Answer 16

The RNA polymerase encounters a specific termination sequence (stop site) -> this releases the RNA strand
The result of transcription is the mRNA precursor (pre-mRNA)

Question 17

U replaces T

Answer 17

In the RNA strand the bases are complementary to the DNA strand, however, Uracil takes the place of Thymine to pair with A

Question 18

E.g.,

Answer 18

DNA -> TAC GCA TGA CTA
RNA -> AUG CGU ACU GAU

Question 19

RNA processing

Answer 19

Pre-mRNA must undergo RNA processing steps:
-Capping at 5’ end
-Polyadenylation at 3’ end
-Splicing

Question 20

Capping at 5’ end

Answer 20

A modified nucleotide 5′ G (cap), is added to the 5’ of the pre-mRNA
▪ Allows the attachment to the ribosome

Question 21

Polyadenylation

Answer 21

A repetition of poly-A (100–250 bp of ribonucleotide A) is added to the 3’ end
▪ increase the stability of mRNA molecules (preventing its degradation)
▪ facilitates its migration to the cytosol

Question 22

Splicing

Answer 22

Not all sequences of a pre-mRNA hold instructions for proteins:
* INTRONS – internal sequences that do not code for proteins
* EXONS – segments that do code for proteins -> expressed

Question 23

Splicing 2

Answer 23

Introns are cut out and exons are joined together
After RNA processing, pre-mRNA becomes a mature mRNA → in cytoplasm

Question 24

Translation - key points

Answer 24

Translation – conversion of the mRNA code (4 nucleotides) into protein (20 amino acids) * In the cytoplasm on the ribosomes -mRNA carries the genetic code in groups of three consecutive nucleotides → Codons * Each codon specifies one amino acid (Universal Genetic Code), except 3 stop codons -Some amino acids are specified by more than one codon