molecular biology - part two

Question 1

What are the three components of a nucleotide?

Answer 1

1. a sugar, which has 5 carbon atoms, so is a pentose sugar
2. a phosphate group, which is the acidic, negatively-charged part of nucleic acids
3. a base that contains nitrogen and has either one or two rings of atoms in its structure

Question 2

What is a nucleic acid? What are two examples of nucleic acids?

Answer 2

• nucleic acids are polymers (a large molecule composed of many repeated subunits) of nucleotides
• DNA and RNA are two nucleic acids

Question 3

How are nucleotides linked together?

Answer 3

• covalent bonds are formed between the phosphate of one nucleotide and the pentose sugar of the next nucleotide, creating a strong backbone for the molecule of alternating sugar/photphate groups, with a base linked to each sugar

Question 4

Differentiate between DNA and RNA

Answer 4

1. The sugar within DNA is deoxyribose whilethe sugar is RNA is ribose; deoxyribose has one fewer oxygen atom than ribose
2. There are usually two polymers of nucleotides in DNA but only one in RNA; polymers are often referred to as strands, so DNA is double-stranded while RNA is single-stranded
3. In DNA, the nucleotide thymine is present while RNA has the nucleotide uracil instead

Question 5

Describe the 3D structure of the DNA molecule

Answer 5

• DNA is a double helix made of two antiparallel strands of nucleotides linked by hydrogen bondings between complementary base pairs
• each strand consists of a chain of nucleotides linked by covalent bonds
• the two strands are parallel but run in opposite directions so they are said to be antiparallel; one strand is oriented in the direction 5’ to 3’ and the other is oriented in the direction 3’ to 5’
• the two strands are wound together to form a double helix
• the strands are held together by hydrogen bonds between the nitrogenous bases; A bonds with T and G bonds with C

Question 6

List the four nitrogenous bases of DNA

Answer 6

1. Adenine
2. Cytosine
3. Guanine
4. Thymine

Question 7

List the four nitrogenous bases of RNA

Answer 7

1. Adenine
2. Cytosine
3. Guanine
4. Uracil

Question 8

What is meant by complementary base pairing?

Answer 8

• adenine is always paired with thymine (or uracil) and they therefore complement each other by forming base pairs
• guanine is always paired with cytosine and they therefore complement each other by forming base pairs
• purines pair up with pyrimidines

Question 9

Pyrimidine

Answer 9

• cytosine, thymine, uracil

- one ring

Question 10

Purine

Answer 10

• guanine, adenine

- two rings fused together

Question 11

Double helix

Answer 11

• used to describe the strcture of DNA
• two strands that wind around each other like a twisted ladder.
• each strand has a backbone made of alternating groups of sugar (deoxyribose) and phosphate groups.

Question 12

How did Crick and Watson discover the structure of DNA?

Answer 12

• The structural organisation of the DNA molecule was correctly proposed in 1953 by James Watson and Francis Crick
• These British scientists constructed models to quickly visualise and assess the viability of potential structures
• Their efforts were guided by an understanding of molecular distances and bond angles developed by Linus Pauling, and were based upon some key experimental discoveries:
  1. DNA is composed of nucleotides made up of a sugar, phosphate and base
  2. DNA is composed of an equal number of purines (A + G) and pyrimidines (C + T)
  3. DNA is organised into a helical structure
• Using trial and error, Watson and Crick were able to assemble a DNA model that demonstrated the following:
  1. DNA strands are antiparallel and form a double helix
  2. DNA strands pair via complementary base pairing (A = T ; C Ξ G)
  3. Outer edges of bases remain exposed (allows access to replicative and transcriptional proteins)
• As Watson and Crick’s model building was based on trial and error, a number of early models possessed faults:
  1. The first model generated was a triple helix
  2. Early models had bases on the outside and sugar-phosphate residues in the centre
  3. Nitrogenous bases were not initially configured correctly and hence did not demonstrate complementarity

Question 13

Explain what is meant by ‘semi-conservative’ in terms of DNA replication

Answer 13

• DNA replication is a semi-conservative process, because when a new double-stranded DNA molecule is formed, it contains one original strand and one newly synthesized strand
• the two strands of the double helix separate, each strand serving as a guide
• new strands are creating by adding nucleotides and linking them together
• because of complementary base pairing, the base sequence on the template strand determines the base sequence of the new strand
• the result is two DNA molecules, both composed of an original strand and a newly synthesized strand, and both identical to the original DNA molecule

Question 14

Outline the three hypotheses that had been proposed for the replication of DNA

Answer 14

1. Conservative Model – An entirely new molecule is synthesised from a DNA template
2. Dispersive Model – New molecules are made of segments of new and old DNA
3. Semi-Conservative Model – Each new molecule consists of one newly synthesised strand and one template strand (this is the currently accepted model)

Question 15

How did Meselson and Stahl’s results support the theory of semi-conservative replication of DNA?

Answer 15

• Meselson and Stahl were able to experimentally test the validity of these three models using radioactive isotopes of nitrogen
• Nitrogen is a key component of DNA and can exist as a heavier 15N or a lighter 14N
• DNA molecules were prepared using the heavier 15N and then induced to replicate in the presence of the lighter 14N
• DNA samples were then separated via centrifugation to determine the composition of DNA in the replicated molecules
• The results after two divisions supported the semi-conservative model of DNA replication
• After one division, DNA molecules were found to contain a mix of 15N and 14N, disproving the conservative model
• After two divisions, some molecules of DNA were found to consist solely of 14N, disproving the dispersive model

Question 16

Differentiate between the leading strand and the lagging strand

Answer 16

When replication begins, the two parent DNA strands are separated. One of these is called the leading strand, and it is replicated continuously in the 3’ to 5’ direction. The other strand is the lagging strand, and it is replicated discontinuously in short sections.

Question 17

Outline the role of helicase

Answer 17

• Helicase unwinds and separates the double-stranded DNA by breaking the hydrogen bonds between base pairs
• This occurs at specific regions (origins of replication), creating a replication fork of two strands running in antiparallel directions
• The two separated polynucleotide strands will act as templates for the synthesis of new complementary strands

Question 18

Outline the role of DNA gyrase

Answer 18

• DNA gyrase reduces the torsional strain created by the unwinding of DNA by helicase
• It does this by relaxing positive supercoils (via negative supercoiling) that would otherwise form during the unwinding of DNA

Question 19

Outline the role of single stranded binding proteins

Answer 19

• SSB proteins bind to the DNA strands after they have been separated and prevent the strands from re-annealing
• These proteins also help to prevent the single stranded DNA from being digested by nucleases
• SSB proteins will be dislodged from the strand when a new complementary strand is synthesised by DNA polymerase III

Question 20

Outline the role of DNA primase

Answer 20

• DNA primase generates a short RNA primer (~10–15 nucleotides) on each of the template strands
• The RNA primer provides an initiation point for DNA polymerase III, which can extend a nucleotide chain but not start one

Question 21

Outline the role of DNA polymerase III

Answer 21

• Free nucleotides align opposite their complementary base partners (A = T ; G = C)
• DNA pol III attaches to the 3’-end of the primer and covalently joins the free nucleotides together in a 5’ → 3’ direction
• As DNA strands are antiparallel, DNA pol III moves in opposite directions on the two strands
• On the leading strand, DNA pol III is moving towards the replication fork and can synthesise continuously
• On the lagging strand, DNA pol III is moving away from the replication fork and synthesises in pieces (Okazaki fragments)

Question 22

Outline the role of DNA polymerase I

Answer 22

• As the lagging strand is synthesised in a series of short fragments, it has multiple RNA primers along its length
• DNA pol I removes the RNA primers from the lagging strand and replaces them with DNA nucleotides

Question 23

Define Okazaki fragment

Answer 23

Okazaki fragments are short, newly synthesized DNA fragments that are formed on the lagging template strand during DNA replication.

Question 24

Outline the role of DNA ligase

Answer 24

• DNA ligase joins the Okazaki fragments together to form a continuous strand
• It does this by covalently joining the sugar-phosphate backbones together with a phosphodiester bond

Question 25

Explain how PCR works and what it does

Answer 25

• polymerase chain reaction
• technique used to make many copies of a selected DNA sequence
• only a very small quantity of the DNA is needed at the start

Steps:

1. DNA is loaded into the PCR machine
2. PCR machine separates DNA strands by heating them up to 95 C for 15 seconds
3. PCR machine quickly cools to 54 C
4. A large amount of primers are present and they bind rapidly to the target sequences, preventing the re-annealing of the parent strands
5. Copying of the single parent strands then starts from the primers
6. Using the single strands with primers as templates, the synthesis of double-stranded DNA begins using the enzyme Taq DNA polymerase
7. PCR machine heats up to 72 C, which is the optimum temperature for Taq DNA polymerase, for 80 seconds
8. PCR machine heats up to 95 degrees again and begins the next cycle

Question 26

Explain why Taq DNA polymerase is used in PCR

Answer 26

• adapted to be very heat-stable to resist denaturation

- can resist the brief 95 C used to separate the DNA strands

Question 27

What are the two processes required to produce a specific polypeptide using the base sequence of a gene?

Answer 27

1. Transcription

2. Translation

Question 28

Define transcription and outline the process

Answer 28

• the synthesis of mRNA copied from the DNA base sequences by RNA polymerase
• RNA is single-stranded, so transcription only occurs along one of the two strands of DNA

Steps:

1. In intiation, RNA polymerase binds to a site (the promoter) on the DNA at the start of a gene
2. In elongation, RNA polymerase moves along the gene in a 5’ to 3’ direction, separating DNA into single strands and pairing up RNA nucleotides with complementary bases on one strand of the DNA; uracil is used in place of thymine when pairing up with adenine
3. RNA polymerase forms covalent bonds between the RNA nucleotides
4. When RNA polymerase reaches the terminator, RNA separates from DNA and double helix reforms
5. Transcription stops at the end of the gene and the completed RNA molecule is released

Question 29

In which direction does transcription occur?

Answer 29

5’ to 3’ direction

Question 30

What is the sense strand?

Answer 30

• the DNA strand with the same base sequence as the RNA

Question 31

What is the antisense strand?

Answer 31

• the DNA strand that acts as the template and has a complementary base sequence to the RNA

Question 32

What is mRNA and what is its function?

Answer 32

• messenger RNA
• carries the information needed to synthesize a polypeptide
• most RNA is mRNA

Question 33

What is tRNA and what is its function?

Answer 33

• transfer RNA

- involved in decoding the base sequences of mRNA into an amino acid sequence during translation

Question 34

What is rRNA and what is its function?

Answer 34

• ribosomal RNA

- part of the structure of ribosome

Question 35

What is a codon?

Answer 35

• three bases on mRNA that correspond to one amino acid in a polypeptide
• differenet codons can code for the same amino acid; for this reason, the code is said to be ‘degenerate’
• there are also ‘stop’ codons that code for the end of translation

Question 36

What is an anticodon?

Answer 36

• three bases on tRNA that is complementary to a codon
• amino acids are carried on tRNA which has a three-base anticodon that is complementary to the mRNA codon for that amino acid

Question 37

What is the start codon?

Answer 37

• AUG (which codes for methionine)

- the first codon of a mRNA transcript translated by a ribosome

Question 38

Deduce the base sequence of the antisense strand transcribed to produce mRNA with the base sequence CUCAUCGAAUAACCC

Answer 38

GAGTAGCTTATTGGG

Question 39

Explain the production of insulin as an example of the universality of the genetic code

Answer 39

• The genetic code is universal – almost every living organism uses the same code (there are a few rare and minor exceptions)
• As the same codons code for the same amino acids in all living things, genetic information is transferrable between species
• The ability to transfer genes between species has been utilised to produce human insulin in bacteria (for mass production)

Steps:

• The gene responsible for insulin production is extracted from a human cell
• It is spliced into a plasmid vector (for autonomous replication and expression) before being inserted into a bacterial cell
• The transgenic bacteria (typically E. coli) are then selected and cultured in a fermentation tank (to increase bacterial numbers)
• The bacteria now produce human insulin, which is harvested, purified and packaged for human use (i.e. by diabetics)

Question 40

Explain the Hershey-Chase experiment and how it provided evidence that DNA is the genetic material

Answer 40

• In the mid-twentieth century, scientists were still unsure as to whether DNA or protein was the genetic material of the cell
• It was known that some viruses consisted solely of DNA and a protein coat and could transfer their genetic material into hosts
• In 1952, Alfred Hershey and Martha Chase conducted a series of experiments to prove that DNA was the genetic material
• Viruses (T2 bacteriophage) were grown in one of two isotopic mediums in order to radioactively label a specific viral component
• Viruses grown in radioactive sulfur (35S) had radiolabelled proteins (sulfur is present in proteins but not DNA)
• Viruses grown in radioactive phosphorus (32P) had radiolabeled DNA (phosphorus is present in DNA but not proteins)
• The viruses were then allowed to infect a bacterium (E. coli) and then the virus and bacteria were separated via centrifugation
• The larger bacteria formed a solid pellet while the smaller viruses remained in the supernatant
• The bacterial pellet was found to be radioactive when infected by the 32P–viruses (DNA) but not the 35S–viruses (protein)
• This demonstrated that DNA, not protein, was the genetic material because DNA was transferred to the bacteria

Question 41

Explain how Rosalind Franklin and Maurice Wilkins investigated DNA structure and their findings

Answer 41

• Rosalind Franklin and Maurice Wilkins used a method of X-ray diffraction to investigate the structure of DNA
• DNA was purified and then fibres were stretched in a thin glass tube (to make most of the strands parallel)
• The DNA was targeted by a X-ray beam, which was diffracted when it contacted an atom
• The scattering pattern of the X-ray was recorded on a film and used to elucidate details of molecular structure
• From the scattering pattern produced by a DNA molecule, certain inferences could be made about its structure:
  1. Composition: DNA is a double stranded molecule
  2. Orientation: Nitrogenous bases are closely packed together on the inside and phosphates form an outer backbone
  3. Shape: The DNA molecule twists at regular intervals (every 34 Angstrom) to form a helix (two strands = double helix)

Question 42

Explain how the structure of DNA suggested a mechanism for DNA replication

Answer 42

1. Franklin’s x-ray diffraction experiments demonstrated that the DNA helix is both tightly packed and regular in structure
   - Phosphates (and sugars) form an outer backbone and nitrogenous bases are packaged within the interior
2. Chargaff had also demonstrated that DNA is composed of an equal number of purines (A + G) and pyrimidines (C + T)
   - This indicates that these nitrogenous bases are paired (purine + pyrimidine) within the double helix
   - In order for this pairing between purines and pyrimidines to occur, the two strands must run in antiparallel directions
3. When Watson & Crick were developing their DNA model, they discovered that an A–T bond was the same length as a G–C bond
   - Adenine and thymine paired via two hydrogen bonds, whereas guanine and cytosine paired via three hydrogen bonds
   - If the bases were always paired this way, then this would describe the regular structure of the DNA helix (shown by Franklin)

Consequently, DNA structure suggests two mechanisms for DNA replication:

• Replication occurs via complementary base pairing (adenine pairs with thymine, guanine pairs with cytosine)
• Replication is bi-directional (proceeds in opposite directions on the two strands) due to the antiparallel nature of the strands

Question 43

What are nucleosomes and what are their roles?

Answer 43

1. In eukaryotic organisms, the DNA is packaged with histone proteins to create a compacted structure called a nucleosome
   - Nucleosomes help to supercoil the DNA, resulting in a greatly compacted structure that allows for more efficient storage
   - Supercoiling helps to protect the DNA from damage and also allows chromosomes to be mobile during mitosis and meiosis
2. Nucleosomes help to regulate transcription in eukaryotes
   - the histone proteins have protruding tails that determine how tightly the DNA is packaged
   - Typically the histone tails have a positive charge and hence associate tightly with the negatively charged DNA
   - Adding an acetyl group to the tail (acetylation) neutralises the charge, making DNA less tightly coiled and increasing transcription
   - Adding a methyl group to the tail (methylation) maintains the positive charge, making DNA more coiled and reducing transcription

Question 44

Explain what is meant by ‘tandem repeats’ and how it can be used for DNA profiling

Answer 44

• DNA profiling is a technique by which individuals can be identified and compared via their respective DNA profiles
• Within the non-coding regions of an individual’s genome there exists satellite DNA – long stretches of DNA made up of repeating elements called short tandem repeats (STRs)
• Tandem repeats can be excised using restriction enzymes and then separated with gel electrophoresis for comparison
• As individuals will likely have different numbers of repeats at a given satellite DNA locus, they will generate unique DNA profiles
• Longer repeats will generate larger fragments, while shorter repeats will generate smaller fragments
• Parernal lineage can be deduced by analyzing short tandem repeats from the Y-chromone and deduce maternal lineage by analyzing mitochondrial DNA variations in single nucleotides at specific locations called hypervariable regions

Question 45

Explain the practical application of using nucleotides containing dideoxyribonucleic acid

Answer 45

• Many copies of the unknown DNA are placed into test tubes with all of the raw materials including nucleotides/enzymes, along with small quantities of dideoxyribonucleotides which have been labelled with different fluorescent markers
• Dideoxynucleotides (ddNTPs) lack the 3’-hydroxyl group necessary for forming a phosphodiester bond
• Consequently, ddNTPs prevent further elongation of a nucleotide chain and effectively terminate replication
• The fragments are separated by length using electrophoresis and the sequence of bases can be automatically analyzed by comparing the colour of fluorescence with the length of the fragment

Question 46

Non-coding regions of DNA

Answer 46

• there are some regions of DNA that do not code for proteins but have other important functions
• most of the eukaryotic genome is non-coding
• includes enhances, silences, promoters, introns

Question 47

Promoter

Answer 47

• non-coding region of DNA with a function
• a sequence that is located near a gene
• it is a binding site of RNA polymerase, the enzyme that catalyses the formation of the covalent bond between nucleotides during the synthesis of RNA
• the promoter itself is not transcribed by plays a role in transcription

Question 48

Enhancer

Answer 48

• non-coding region of DNA with a function
• a sequence to which proteins can bind to regulate transcription
• increase the likelihood that transcription of a particular gene will occur

Question 49

Silencer

Answer 49

• non-coding region of DNA with a function
• a sequence to which proteins can bind to regulate transcription
• induces a negative effect on the transcription of its particular gene

Question 50

Explain how proteins regulate gene expression using lactose as an example

Answer 50

• Activator proteins bind to enhancer sites and increase the rate of transcription (by mediating complex formation)
• Repressor proteins bind to silencer sequences and decrease the rate of transcription (by preventing complex formation)

Lactose (can also be used as an example of how chemical environment impacts gene expression)

• the genes responsible for the absorption/metabolsim of lactose are expressed in the presence of lactose and are not expressed in the absense of lactose
• in the presense of lactose, a repressor protein is deactivtated
• once the lactose has been broken down, the repressor protein is re-activated and proceeds to block the expression of lactose metabolism genes

Question 51

Explain how the environment of a cell/organism impacts gene expression

Answer 51

• production of skin pigmentation during exposure to sunlight in humans
• in embryonic development, the embryo contains an uneven distribution of chemicals called morphogens
• concentrations of the morphogens affect gene expression, contributing to different patterns of gene expression and thus different fates of the embryonic cells depending on their position in the embryo
• in coat colour of Siamese cats, a certain mutation allows normal pigment porduction only at temperatures below body temperature; at higher temperatures, the protein product is inactive or less active, resulting in less pigment
• consumption of products containing lactose will deactivate a repressor protein, allowing the genes for lactose metabolism to be expressed

Question 52

Explain the significance of the modification of histone tails

Answer 52

• chemical modification of histone tails is an important factor in determining whether a gene will be expressed or not
• a number of different types of modfiication can occur to the histone tails, including the addition of an acetyl group, the addition of a methyl group or the addition of a phosphate group
• adding an acetyl group to the tail (acetylation) neutralises the charge, making DNA less tightly coiled and increasing transcription
• adding a methyl group to the tail (methylation) maintains the positive charge, making DNA more coiled and reducing transcription
• thus, chemical modification of histone tails can either activate or deactivate genes by decreasing or increasing the accesibility of the gene to transcription factors

Question 53

Explain the impact of DNA methylation

Answer 53

• Direct methylation of DNA (as opposed to the histone tails) can also affect gene expression patterns
• Increased methylation of DNA decreases gene expression (by preventing the binding of transcription factors)
• Consequently, genes that are not transcribed tend to exhibit more DNA methylation than genes that are actively transcribed
• The amount of DNA methylation varies during a lifetime and is affected by environmental factors

Question 54

What is meant by pre-mRNA and mature mRNA?

Answer 54

• the immediate product of mRNA trancription is referred to as pre-mRNA
• after it goes through several stages of post-transcriptional modification, it is referred to as mature mRNA

Question 55

Explain post-transcriptional modification and its significance

Answer 55

• In eukaryotes, there are three post-transcriptional events that must occur in order to form mature messenger RNA:

1. Capping
   - Capping involves the addition of a methyl group to the 5’-end of the transcribed RNA
   - The methylated cap provides protection against degradation by exonucleases
   - It also allows the transcript to be recognised by the cell’s translational machinery (e.g. nuclear export proteins and ribosome)
2. Polyadenylation
   - Polyadenylation describes the addition of a long chain of adenine nucleotides (a poly-A tail) to the 3’-end of the transcript
   - The poly-A tail improves the stability of the RNA transcript and facilitates its export from the nucleus
3. Splicing
   - Within eukaryotic genes are non-coding sequences called introns, which must be removed prior to forming mature mRNA
   - The coding regions are called exons and these are fused together when introns are removed to form a continuous sequence
   - Introns are intruding sequences whereas exons are expressing sequences
   - The process by which introns are removed is called splicing

Question 56

Intron

Answer 56

• sequences that will not contribute to the formation of the polypeptide
• interspersed throughout the pre-mRNA and are removed during post-transcriptional modification

Question 57

Exon

Answer 57

• the coding portion of the pre-mRNA that will be spliced togethed to form the mature mRNA
• will contribute to the formation of the polypeptide

Question 58

Explain how tRNA-activating enzymes illustrate enzyme-substrate specificity and the role of phosphorylation

Answer 58

• Each tRNA molecule binds with a specific amino acid in the cytoplasm in a reaction catalysed by a tRNA-activating enzyme
• Each amino acid is recognised by a specific enzyme (the enzyme may recognise multiple tRNA molecules due to degeneracy)
• The binding of an amino acid to the tRNA acceptor stem occurs as a result of a two-step process:
• The enzyme binds ATP to the amino acid to form an amino acid–AMP complex linked by a high energy bond (PP released)
• The amino acid is then coupled to tRNA and the AMP is released – the tRNA molecule is now “charged” and ready for use
• The function of the ATP (phosphorylation) is to create a high energy bond that is transferred to the tRNA molecule
• This stored energy will provide the majority of the energy required for peptide bond formation during translation

Question 59

How is translation intiated?

Answer 59

The first stage of translation involves the assembly of the three components that carry out the process (mRNA, tRNA, ribosome)

• The small ribosomal subunit binds to the 5’-end of the mRNA and moves along it until it reaches the start codon (AUG)
• Next, the appropriate tRNA molecule bind to the codon via its anticodon (according to complementary base pairing)
• Finally, the large ribosomal subunit aligns itself to the tRNA molecule at the P site and forms a complex with the small subunit

Question 60

Explain the elongation stage of translation

Answer 60

• A second tRNA molecule pairs with the next codon in the ribosomal A site
• The amino acid in the P site is covalently attached via a peptide bond (condensation reaction) to the amino acid in the A site
• The tRNA in the P site is now deacylated (no amino acid), while the tRNA in the A site carries the peptide chain
• The ribosome moves along the mRNA strand by one codon position (in a 5’ → 3’ direction)
• The deacylated tRNA moves into the E site and is released, while the tRNA carrying the peptide chain moves to the P site
• Another tRNA molecules attaches to the next codon in the now unoccupied A site and the process is repeated

Question 61

How is translation terminated?

Answer 61

The final stage of translation involves the disassembly of the components and the release of a polypeptide chain

• Elongation and translocation continue in a repeating cycle until the ribosome reaches a stop codon
• These codons do not recruit a tRNA molecule, but instead recruit a release factor that signals for translation to stop
• The polypeptide is released and the ribosome disassembles back into its two independent subunits

Question 62

Differentiate between the role of free ribosomes and bound ribosomes

Answer 62

• in eukaryotes, proteins function in a particular cellular compartment
• proteins are synthesized either in the cytoplasm or at the ER depending on the final destiination of the protein
• free ribosomes synthesize proteins for use primarily within the cell (ie. cytoplasm, mitochondria, chloroplasts)
• bound ribosomes synthesize proteins primarily for secretion or for use in lysosomes/ER/Golgi apparatus
• whether the ribosome is free or bound to the ER depends on the presence of a signal sequence on the polypeptide being translated

Question 63

Why is there a delay between transcription/translation in eukaryotes but not in prokaryotes?

Answer 63

• translation can occur immediately after transcription in prokaryotes due to the absence of a nuclear membrane
• in eukaryotes, cellular functions are compartmentalized whereas in prokaryotes, they are not
• once transcription is complete in eukaryotes, the transcript is modified in several ways before exiting the nucleus
• there is thus a delay between transcription and translation due to compartmentalization
• in prokaryotes, as soon as the mRNA is transcribed, translation begins

Question 64

Primary structure of protein

Answer 64

• The order and number of the amino acid sequence in polypeptides is called the primary structure and determines the way the chain will fold
• Different amino acid sequences will fold into different configurations due to the chemical properties of the variable side chains

Question 65

Secondaty structure of protein

Answer 65

• Amino acid sequences will commonly fold into two stable configurations, called secondary structures
• The secondary structure is the formation of alpha helices and beta pleated sheets stabilized by hydrogen bonding
• There is a tendency for a carboxyl group (C=O) in one part of the chain and an amino group (N-H) in another part of the chain to form a hydrogen bond
• Alpha helices occur when the amino acid sequence folds into a coil / spiral arrangement
• Beta-pleated sheets occur when the amino acid sequence adopts a directionally-oriented staggered strand conformation

Question 66

Tertiary structure of protein

Answer 66

• The overall three-dimensional configuration of the protein is referred to as the tertiary structure of the protein
• The tertiary structure is the further folding of the polypeptide stabilized by interactions between R groups
• There are several different types of interaction:
  1. Positively charged R-groups will interact with negatively charged R-groups
  2. Hydrophobic amino acids will orientate themselves toward the centre of the polypeptide to avoid contact with water, while hydrophilic amino acids will orientate themselves outwards
  3. Polar R-groups will form hydrogen bonds with other polar R-groups
  4. The R-group of the amino acid cysteine can form a covalent bond with the R-group of another cysteine forming what is called a disulphide bridge

Question 67

Quatenary structure of protein

Answer 67

• exists in proteins with more than one polypeptide chain
• proteins can be formed from a single polypeptide chain or from multiple polypeptide chains
• quaternary structure refers to the way polypeptides fit together when there is more than one chain; it also refers to the addition of non-polypeptide components