Proteins & Recombinant DNA

Question 1

Recombinant DNA

Answer 1

A molecule of DNA that has originated from 2 or more DNA fragments that aren’t found together in nature.

Question 2

Cloning

Answer 2

The production of identical copies of a particular DNA molecule. The isolation of a particular piece of DNA from the rest of a cells DNA.

Question 3

Steps in Cloning a DNA Fragment

Answer 3

1. A plasmid vector and the specific DNA fragment that you want cloned are combined. The DNA fragment is enzymatically inserted into the plasmid which creates a recombinant plasmid.
2. E.coli cells are added and the plasmids will then enter these cells.
3. The plasmid DNA that is now inside the E.coli cells will replicate.
4. The bacteria are all placed in a solution with a buffer which has a substance which the cells with the inserted DNA are immune to however any cells that didn’t uptake the plasmids will die in order to leave behind only the chosen sample of cells.
5. The E.coli cells that survived will also multiply meaning that more bacterial cells with the desired plasmid are formed.

Question 4

Cloning DNA with Restriction Enzymes

Answer 4

The cloning process can be used to make many copies of a DNA fragment to produce proteins in bacteria or to study gene mutation. Plasmids and DNA that are to be combined into a new plasmid can be cut using a restriction enzyme. This produces complimentary ends with sequence specific restriction enzymes. To mix these the plasmid is cut, the DNA fragment is cut and put together, DNA ligase combines the 2 using ATP. The result is a new plasmid with the desired DNA fragment contained within it.

Question 5

Restriction Enzymes

Answer 5

These are endonucleases (endo DNases). These digest double-stranded DNA at internal phosphodiester bonds. They cut DNA at specific sites which are defined by nucleotide sequences also known as recognition sequence, restriction sites or restriction sequences. These areas are palindromes (are the same sequence when read normally or backward) and have 4-8 nucleotides (some may have more). These leave the 3’-hydroxul and 5’-phosphate groups on both cut strands.

Question 6

Palindromic DNA Sequences

Answer 6

These are the recognition sites of the restriction enzymes and the specific pattern will have a specific restriction enzyme that can bind to it.

Question 7

Restriction Enzyme Cuts

Answer 7

These can produce blunt ends where both of the DNA strands are cut off at the same position e.g. Hpal. They can produce sticky or cohesive ends where the cut will leave a 5’ overhang (the longer strand goes from 3’-5’) e.g. EcoRI or a 3’ overhang (the longer strand goes from 5’-3’) e.g. Pstl.

Question 8

Naming Restriction Enzymes

Answer 8

The first letter is from the Genus e.g. (EcoRI = Escherichia) and the next 2 are from the Species (EcoRI = coli). Another example is Hemophilus influenzae = HindIII. The final part of the name refers to the type of enzyme and the number e.g. (EcoRI = restriction enzyme 1). Another convention of naming is that the first 3 letters should be italicized and the rest shouldn’t be.

Question 9

Annealing DNA

Answer 9

For sticky ends if the 2 overhanging sections of DNA match up they will join together via base pairing and a DNA ligase will make up the remainder of the phosphate and hydroxyl groups that are missing. For blunt ends they can’t base pair however can be attached via ligase however the probability of this occurring is much lower than the joining rate for sticky ends.

Question 10

Where Restriction Enzymes Come From

Answer 10

Initially these came about as a protective asset for bacteria against viruses. If a bacteria is infected by the viral DNA and will take over the replication machinery of the bacteria and produce more viral components. In order to protect from the viruses bacteria will methylate the DNA at certain spots in order to hide the recognition sites from restriction enzymes which stops the viral DNA from replicating. The viral DNA however isn’t methylated and so will be destroyed by the restriction enzyme.

Question 11

DNA Ligase

Answer 11

This enzyme covalently links the 3’ hydroxyl and 5’ phosphate groups using ATP as an energy source. The DNA is then joined which is known as ligated.

Question 12

Plasmids

Answer 12

These are small, extrachromosomal, double-stranded, circular DNA molecules which are found in bacterial cells. These carry genes that are beneficial for the survival of the bacteria under certain conditions e.g. antibiotic resistance. This is distinct from the bacterial chromosome and replicate independently from the chromosomal DNA. They are also known as ‘Vectors as they move DNA from one place to another (they can transfer into other bacteria).

Question 13

Features of Plasmids

Answer 13

They have promoters which allows for the expression of a gene cloned downstream of the promoter (where RNA is started to be made). They are the sites for restriction enzymes which allows for the cloning of particular sections of DNA, typically there are many unique restriction enzyme sites which allows for multiple cloning (also known as ‘multiple cloning site’ or ‘polylinker’). They are the origin of replication which allows them to be passed onto daughter cells during division with several copies being present in a single cell. They are a selectable marker as they can be resistant or vulnerable to certain things due to the presence of the specific DNA contained.

Question 14

Transformation of Bacterial Cells

Answer 14

There are 2 main methods of heat shock or electroporation. In heat shock the bacteria are grown at 37C and are heated up to 42C which opens up the cell wall and the membrane becomes a bit looser which allows DNA to enter the cell. In electroporation a current (high voltage for a short period of time) is applied to the cells which makes the bacterial cell wall and membrane more permeable to DNA.

Question 15

Agarose Gels

Answer 15

Agarose is a red algal (from algae) carbohydrate. In order to produce this you boil it in buffer to dissolve it and pour into a tray (once cooled to 60C) containing a comb which leads to a solid gel with loading pockets forming when it cools down. The combs can be pulled out and the tray with gel can be transferred to an electrophoresis tank and overlay with a buffer.

Question 16

Agarose Gel Electrophoresis

Answer 16

First a DNA molecule is digested into fragments by a restriction enzyme and then undergoes gel electrophoresis. In this process the gel is placed in an electrophoresis chamber which has a positive and negative electrode. A voltage is applied to the DNA which is added to the loading pockets (near the negative electrode) and due to the negative charge of the DNA it migrates to the positive electrode. The DNA moves based on its size (molecular weight) with the smallest moving the fastest (and therefore furthest). This leaves the plate with DNA fragments separated by size.

Question 17

Large Scale Production of Proteins

Answer 17

Once the plasmid is entered into the host cell (bacteria, yeast, mammal cells, insect cells) a new protein will be produced. There is a promoter sequence present in the plasmid DNA which allows for the mRNA to produce the protein of the inserted DNA. There is also a tag that is behind the DNA sequence which allows for the purification of the protein from the other proteins in the cell. This is done via affinity chromatography when a binding molecule is added which is chemically inert and binds to the proteins with a particular tag while other proteins will simply pass by. The non-bound proteins are washed away which leaves the remaining purified protein.

Question 18

Obtaining DNA Fragments

Answer 18

The DNA sequence is typically known through genome sequences or published in certain data bases. If you don’t want the introns you can translate the mRNA into DNA code for the specific gene and send it to a company which will send you samples of that particular sequence. Some genomes aren’t completely known. If part or all of the DNA is known you can synthesise the required DNA fragment using polymerised chain reaction (PCR).

Question 19

Polymerised Chain Reaction (PCR)

Answer 19

DNA polymerase releases a pyrophosphate (diphosphate = 2 phosphate molecules) from dNTPs (deoxyribonucleoside triphosphates) and adds the resulting dNMP (deoxyribonucleoside monophosphate) to the 3’-hydroxyl end of the primer strand via a nucleophile attack mechanism. The pyrophosphate is cleaved by pyrophosphatase in a non-reversible reaction. The new strand of DNA is synthesised in the 5’-3’ direction. The template strand is read in the 3’-5’ direction. In this process a heat stable DNA polymerase (Taq DNA Polymerase) from thermophilic bacteria (living in hot springs) is used. The proteins required for primers and DNA separation aren’t required in this process as heat is used as a substitute.

Question 20

PCR Requirements

Answer 20

A template DNA strand to be amplified is required. 2 Primers which are short, single-stranded DNA (ssDNA) oligonucleotides that can bind to either the upper or the lower template. Taq DNA polymerase is the enzyme that synthesises copies of the template DNA. dNTPs (dATP, dTTP, dCTP, dGTP) which are the building blocks for the newly synthesised DNA. A buffer ensures that the reaction conditions are suitable for the amplification. Magnesium ions (Mg2+) are required as a cofactor for the Taq DNA polymerase (the enzyme won’t work without it). A PCR machine that provides the correct temperatures and timing for individual PCR steps.

Question 21

Primers for PCR

Answer 21

The sequence information is required to design these. These must be designed so they can base pair with the template DNA strand. 2 of these are required to be attached to either strand of DNA. These are typically 20-30 nucleotides in length. These can be ordered from the companies which specialises in this process.

Question 22

PCR Cycle

Answer 22

In the first cycle 2 double stranded DNA molecules are created. The DNA to be amplified is separated and the oligonucleotide primers are added. The Taq DNA polymerase will then attaches to each of the primers and synthesises the DNA for both strands. In the second cycle 4 double stranded DNA molecules are created. The 2 double stranded DNA molecules from the first cycle are then separated into single strands with primers attached to them. The Taq DNA polymerase will then begin to synthesise new DNA molecules resulting in 4 double stranded DNA molecules. This cycle will continue to occur for 20-40 cycles until an adequate amount of DNA fragments is produced.

Question 23

Duplication Power

Answer 23

After 1 cycle there is 2 DNA molecules, after 2 cycles there is 4, after 3 cycles there is 8. The mathematical expression is 2^n (n = number of cycles). After 30 cycles there will be 1,073,741,824 DNA molecules.

Question 24

PCR Temperatures

Answer 24

In order to separate the 2 DNA strands requires 95C of heat which breaks the non-covalent hydrogen bonds between the base pairs. In order for the primers to anneal (bind) to the single strands the temperature is 50-60C depending on the size of the primer size (calculated based on the number of G and C bases as they have 3 hydrogen bonds). In order to synthesise the new strand of DNA the temperature is set to 72C which is the ideal for Taq DNA polymerase (durability and efficiency).

Question 25

Thermocycler

Answer 25

The first PCR was invented by Kary Mullis in 1983 and received a Nobel Prize in 1993. The first time was done at 37C as the Taq DNA polymerase wasn’t yet discovered and later it was done at higher temperature however he would stand by the reaction and continue to add DNA polymerase as they wouldn’t survive in the high temperatures. With modern machines 96 microtiter plates in which 96 PCR reaction can occur simultaneously in 2.5 hours with temperature resistant DNA polymerase meaning it is completely automatic.

Question 26

Application of PCR

Answer 26

This process can be used to clone and analyse DNA. It can help to genotype and classify organisms. DNA fingerprinting as molecular markers of disease or crime suspects or parental studies. Mutational screening to identify the mutations related to a trait or some type of disease. Site directed mutagenesis to create mutation and study the impact of them. Detection of pathogens to find a pathogen by their DNA marker. Molecular epidemiology to follow a sequence indicative of a pathogen through a population. Pre-natal diagnostics to identify potential risk factors. Molecular archaeology to clone and sequence DNA from a fossil to relate to a current species. Molecular ecology to identify risk factors in the environment. Gene expression studies.

Question 27

Cloning a DNA Fragment

Answer 27

Once a DNA fragment has been cloned all of the restriction sites expect 2 should be removed and then the fragments can be combined. They can be ligated into a plasmid vector.

Question 28

Parental Markers

Answer 28

A molecular marker is any site (locus) in the genome of an organism at which the DNA base sequence varies among the different individuals of a population. In finding biological parents the DNA can be ran through a gel electrophoresis and the movements of the DNA molecules can be compared to see whether a child is related to a particular parent or not.

Question 29

Crime Scene Investigations

Answer 29

If some small evidence is found at the site of the crime you can use PCR to replicate the DNA found and then run in through a gel electrophoresis. Once suspects for the case are found their DNA can also undergo this process for a gel electrophoresis to see who was there at the scene of the crime.

Question 30

Environmental Studies

Answer 30

In soil a bacterium called Rhizobium leguminosarum is found. In healthier soils this bacteria is found in many different varieties however in solid which are contaminated by heavy metals there are only a few types of this bacteria. In order to analyse whether a soil is contaminated or not you could run the soil samples through a PCR to amplify the amount of bacterial DNA and then through a gel electrophoresis and if the line of DNA is slightly skewed meaning they move different amount then many varieties of bacterium are present meaning uncontaminated soil however if all the lines are in a straight line only 1 type of the bacterial DNA is present and the soil is contaminated.

Question 31

Reverse Transcription-Polymerase Chain Reaction (RT-PCR)

Answer 31

In this process mRNA is cloned to form DNA. First the mRNA must be isolated to be cloned. Once this occurs the first prime, reverse transcriptase and deoxyribonucleoside triphosphates (dNTPs) are added. The reaction then occurs where a single strand of DNA is attached to the mRNA. This newly formed RRNA:DNA hybrid molecule has the RNA removed and a new DNA strand added so it can undergo standard PCR and create cloned DNA. This occurs naturally in viruses which only have RNA and therefore create DNA in order to infect the host cell and create the viral proteins it requires.

Question 32

Reverse Transcription (RT) of RNA

Answer 32

All eukaryotic mRNA have a poly-A tail, and hybridize with a poly-T primer. Reverse transcriptase is an RNA dependent DNA polymerase found in retroviruses which requires a primer. It synthesises DNA in the 5’3’ direction creating a DNA copy of mRNA which creates an RNA:DNA hybrid. RNaseH (H = hybrid) degrades the RNA in this hybrid molecule however leaves some of it behind as a primer for complementary DNA (cDNA) strand synthesis. DNA polymerase will then synthesise the complementary DNA strand which is then amplified (cloned) through vectors and bacteria or using PCR.

Question 33

RT-PCR in Gene Expression Studies

Answer 33

Studies involving the evolutionary origins of mammalian immunoglobulin heavy chain isotypes in different antibody types (igM, IgD, IgG, IgE and IgA). As part of this study researchers investigated tissue specific mRNA expression of the different immunoglobulins and found that IgO is only expressed in the spleen. This process was used to detect the Ig gene expression in different tissues of a platypus. Researcher weren’t able to conduct Northern blotting to detect tissue expression of the platypus Ig genes due to the unavailability of high-quality RNA. This is why they used this specific procedure.

Question 34

Dideoxyribonucleotide Triphosphates (ddNTPs)

Answer 34

This nucleotide molecule also lacks a hydroxyl group at carbon 2’ similar to a dNTP however unlike the dNTP it also lacks a hydroxyl group at carbon 3’. Once this molecule is added to a DNA chain it causes termination as the lack of a hydroxyl group on carbon 3’ means it can’t attach to another nucleotide so it can’t be elongated.

Question 35

Sanger DNA Sequencing Method

Answer 35

This requires the use of ddNTPs in combination of dNTPs. The mixture of dNTPs, primers, template DNA, DNA polymerase and 1/4 ddNTPs (ddATP, ddTTP, ddCTP, ddGTP) were separated into 4 separate tubes. The ddNTP determines at which nucleotide a chain termination occurs and the dNTP:ddNTP determines the length of the synthesised chain. These 4 reaction tubes made multiple lengths of DNA molecules which were terminated by a ddNTP molecule. All 4 tubes were ran through gel electrophoresis to separate the DNA fragments by size. After this is done the ddNTP molecules nitrogenous base can be read in order from shortest to longest molecule (furthest to closest travelled) to figure out the original template DNA sequence. In order to visualise the DNA they were labelled with radioactive phosphate which would be detected using an X-ray film.

Question 36

Modern Sequencing

Answer 36

This still uses many of the principle of the Sanger method however there are some changes. There are some changes with the use of different fluorescent labels (not radiation due to contamination and cleanup). The DNA synthesis is done in the presence of all 4 dNTPs and all 4 labelled ddNTPs in 1 tube. The ddNTPs still cause termination of DNA and the fragments are ran through capillary gel electrophoresis with the same separation principles (smaller = faster travel and vise versa). A laser is used to excite the fluorescent labels which emit a specific light that can be detected by a computer over time (as it passes through the capillary gel electrophoresis) to produce a chromatogram (helps to identify the order of nitrogenous bases). This method is most efficient for measuring strands of unknown DNA but isn’t used for entire genome sequences.

Question 37

Identifying DNA Fragments in Simple Genomes

Answer 37

Using restriction enzymes to digest a DNA molecule and undertaking a gel electrophoresis a map of simple DNA genomes can be generated. If the DNA sequence is known the map can be used as a reference to determine what the fragments of the sequence are depending on how far it travelled which correlates to how large it is.

Question 38

Identifying DNA Fragments in Complex Genomes

Answer 38

When the DNA genome is complex the restriction enzymes will create a digest with thousands of fragments differing by size. When the gel electrophoresis is done it appears as a smear to the naked eye however if the gel resolution and zoom ability was possible you could differentiate fragments with 1-2 nucleotide length differences. In order to locate a specific fragment within the smear a labelled probe must be hybridised (pairing a dsDNA molecule to the ssDNA fragment) to the DNA fragment.

Question 39

Hybridisation Examples

Answer 39

This process is based on the denaturation and renaturation of DNA. In a denaturation excess heat/pH can split the dsDNA by breaking the hydrogen bonds between nitrogenous bases. In a renaturation heat/pH decrease can restore the dsDNA as the hydrogen bonds are reformed. This process also occurs when complementary RNA strands form dsRNA molecules and when complementary DNA and RNA strands form RNA/DNA hybrids.

Question 40

Hybridisation for DNA Fragment Identification

Answer 40

This is done between a probe and a DNA strand. The probes are ssDNA,15-1000 nucleotides long, attaching to ssDNA molecules and the results of the attachment are dependent on temperature. Formamide lower the melting temperature of nucleic acid duplexes (facilitates the formation of complementary strands). The homologous probes allow us to identify the exact DNA fragment we require whereas the heterologous probes allow us to identify closely related (same gene family) DNA fragments as well which may serve similar purposes if tested.

Question 41

Probes

Answer 41

With homologous versions they detect identical nucleic acid molecules (100% complementary) at higher temperatures meaning only perfect base pairing. With heterologous versions detect related nucleic acid molecules (not necessarily 100% complementary) at lower temperatures meaning imperfect base pairing is possible.

Question 42

Hybridisation Techniques

Answer 42

These methods are used to detect specific DNA and RNA molecules. For DNA Southern blotting is used which is gel based and uses a blot in combination with this method. Fluorescent in situ hybridisation (FISH) is another process used for DNA fragment detection which is cellular based and this method is done using tissue selections. For RNA northern blotting is used with the same requirements as Southern blotting as well as in situ hybridisation which is the same as the FISH method just for RNA.

Question 43

Southern Blotting

Answer 43

In this process the DNA is isolated from cells. The DNA is typically digested with restriction enzymes and then is separated using a 0.5-2.0% agarose gel. DNA is then denatured in alkali solution which creates ssDNA which is transferred overnight using alkali solution or a buffer. The buffer carries DNA from the gel to the membrane (nitrocellulose or nylon) whereas the buffer goes through into the paper towels stacked atop the membrane. Once this is done the DNA is washed (removes gel) and sealed in a bag with probes to hybridise in a heated environment with the DNA which reveals the positions of labelled markers. This can be excited in order to remove the DNA for use.

Question 44

Northern Blotting

Answer 44

In this process the RNA is isolated from cells. The RNA is separated using a 1-1.5% gel in the presence of a denaturing agent which removes any secondary structures from the RNA. The RNA is then transferred overnight using buffer which carries the RNA from the gel to the membrane. The RNA stays on the membrane while the buffer is absorbed by paper towels. Once this is done the RNA is washed (removes gel) and sealed in a bag with probes to hybridise in a heated environment with the RNA which reveals the positions of labelled markers. This can be excited in order to remove the RNA for use.

Question 45

Autoradiogram

Answer 45

After Southern blotting this representation allows us to identify related genes, determine the number of related genes, determines the sizes of related genes and construct restriction maps based on the information. After northern blotting this representation allows us to identify related transcripts, the number of related transcripts, determine the sizes of related transcripts, determine patterns of gene expression e.g. tissues or cell types and development.

Question 46

Autoradiogram of a Northern Blot with Radioactive Labelled Probe

Answer 46

A gene (beta actin) is expressed constitutive meaning at the same level of expression in all those cell lines which controls the amount of RNA being loaded into each lane of the gel electrophoresis. BAG-1 expression changes in relation to beta actin expression in a cell line dependent manner. There was a differential expression of anti-apoptotic (prevents cell death) gene BAG-1 in normal and breast cancer cell lines in people. This shows that this gene could be part of the reason why breast cancer cells don’t die as easily as it showed very little expression in normal cells however certain cancer cells showed higher expression of the gene while some cancer cells didn’t.

Question 47

Making a Probe

Answer 47

DNA is denatured to generate ssDNA that can act as templates for DNA synthesis. Hexanucleotides (DNA primers which are 6 nucleotides long) which are varied in the combinations of bases increasing the chance a DNA molecule can be labelled. DNA polymerase adds nucleotides to the primer to synthesis complementary DNA strands. As some of the nucleotides used were labelled it results in a population of DNA molecules that contain labelled examples of all sequences of both DNA strands. There are 2 types of labels typically used which include a radioactive labelled dNTP or a chemically labelled dNTP.

Question 48

Types of Labelling

Answer 48

In radioactive forms one of the phosphates in dATP is the radioactive 32 isotopic form which can be detected by exposure to x-ray film. The radioactive isotope is introduced to each dATP added to the DNA strand following the probe. In chemical forms a dTTP molecule is marked by digoxygenin (DIG) which is attached with a spacer (a chain which separates the chemical and nucleotide). anti-digoxygenin antibodies will then detect the marked DNA. To detect this chemical the antibodies that detect it are covalently bonded to enzymes. When the antibodies are added a colourless substrate is added which enzymes will turn into a coloured (usually red) substrate which is used to identify the DNA fragments using a photo or other imaging techniques.

Question 49

Fluorescent In Situ Hybridisation (FISH)

Answer 49

This process is used for the approximate mapping of genes directly on chromosomes.
1. Cells and nuclei are broken open.
2. Chromosomes are spread on a microscope slide.
3. The chromosome are then denatured.
4. The fluorescent labelled DNA probe is hybridised to the chromosomes.
5. This is visualised under a fluorescent microscope which gives the approximate location of a target gene.
Multiple labels with different fluorescence allows for the mapping of many genes to chromosomes called chromosome painting which is used for karyotype analysis.

Question 50

In Situ Hybridisation

Answer 50

This process is used to localise transcripts within tissues/cells to determine expression patterns of specific genes in tissues e.g. during development. Using a sense probe the sequence is the same as the transcript meaning hybridisation doesn’t occur which acts as a negative control. Using an antisense probe the sequence is complementary to the transcript and hybridisation occurs to localise transcripts with tissues/cells.

Question 51

Proteins

Answer 51

These are the ultimate end of gene expression, they are abundant in all cells, they are the most diverse group of molecules in the cell (prokaryotes and eukaryotes) and are involved in nearly all cellular process, they do cellular work and provide structures.

Question 52

Amino Acids

Answer 52

The are made up of an amino group (NH2), a carboxyl group (COOH), an alpha (central) carbon and a side chain. These can be ionised when the amino group gains a hydrogen (NH3+) and the carboxyl group loses a hydrogen (COO-). The alpha carbon is asymmetric which allows for 2 stereoisomers (mirror images) to be formed which are an L (amino group on left and carboxyl group on right) and D (amino group on right and carboxyl group on left) stereoisomers. Proteins contain L stereoisomers. Glycine is an exception as it doesn’t have a side chain only a hydrogen meaning there aren’t 4 distinct groups (2x hydrogen groups) on the alpha carbon meaning a stereoisomer can’t exist.

Question 53

Amino Acid Side Chains

Answer 53

These parts of the amino acid differs between all different amino acids which gives each of the amino acids their different characteristics. These characteristics are reflected in the protein structure and function.

Question 54

Amino Acid Charge

Answer 54

At a neutral pH both the amino and carboxyl groups are charged and therefore the charge of the amino acid is neutral. In conditions of low pH the amino acid is positively charged as there is an excess of H+ which means the carboxyl group won’t lose is hydrogen and the amino group will gain one. In conditions of high pH the amino acid is negatively charged as there is a lack of H+ causing the carboxyl group to lose its hydrogen while the amino group doesn’t gain one. All of these conditions are dependent on a non-charged side chain which if it were charged could cause a different result.

Question 55

Protein Charges

Answer 55

In these structures the charge is controlled mainly by the charges of the side chains of the amino acids. The N-terminus and C-terminus are also charged so if all side-chains are neutral then a difference in pH could also change the charge of the protein.

Question 56

Charged Amino Acids

Answer 56

There are only 5 of these. Aspartic acid (Asp) and glutamic acid (Glu) are the only negatively charged ones. Arginine (Arg), Lysine (Lys) and Histidine (His) are the only positively charged ones.

Question 57

Non-Polar Amino Acids

Answer 57

This consists of Gly, Ala, Val, Leu, Ile, Phe, Trp, Pro, Met and Cys. These all have non-polar side chains meaning the amino acids are also non-polar. Met and Cys both also contain a Sulphur group with Cys containing a thiol group (SH). Pro is the only amino acid that is a secondary amine as the amino group is bonded to by 2 different carbons is a 5 molecule ring all the rest being primary amines with only 1 carbon joined to the amino group.

Question 58

Uncharged Polar Amino Acids

Answer 58

This consists of Ser, Thr, Tyr, Asn and Gln. These all either contain an extra amino group or a hydroxyl group. Ser, Thr and Try all contain an extra hydroxyl group to make them polar whereas Asn and Gln contain an extra amino group to make them polar.

Question 59

Acidic & Basic Amino Acids

Answer 59

This consists of Asp, Glu, Lys, His and Arg. These all either contain a side chain that is positive at pH 7 or negative at pH7. Asp and Glu both contain an acidic side chain with an extra carboxylic acid group on the end which loses a H+ and becomes negatively charged at pH 7. Lys, His and Arg all contain a basic side chain with extra amine groups which can gain a H+ and become positively charges at pH7. His has 2 extra amine groups in a 5 molecule ring meaning that both of those parts are secondary amines.

Question 60

Peptide Bonds

Answer 60

The formations of bonds between amino acids through condensation reactions which form peptides. An oligopeptide is around 10 amino acids in length whereas a polypeptide is equal to or greater than 10 amino acids in length.

Question 61

Polypeptides

Answer 61

Due to the resonance of the amide structure the bond between the carbonyl carbon and the nitrogen have a partial double bond character meaning that there is no rotation in peptide bonds. This resonance is caused by the fluctuation of the double bond between the carbon-oxygen bond and the carbon-nitrogen bond which creates a rigid, planar structure. This lack of rotation also impacts the 3D peptide structure. All other single bonds in either amino acid can rotate however.

Question 62

Polypeptide Flexibility

Answer 62

Due to the structure of the side chains of the amino acids there is room for movements within the protein. Due to the size, locations and number of side chains they can form non-covalent interactions with one another which can result in structural changes in the protein. Van der Waals interaction, hydrogen bonds and electrostatic (ionic) interactions can occur. These interactions are what allow proteins to fold into functional structures. If this folding it is observed that the polar side chains will be found on the outside of the folded structure and the non-polar side chains within the core region of the protein which is due to the water content of the cytoplasm which attracts the hydrophilic polar molecules and repels the hydrophobic non-polar side chains.

Question 63

Disulfide Bonds

Answer 63

This is the only type of covalent bond created between 2 side chains of an amino acid. It can only occur between the thiol groups found on 2 Cys amino acid side chains. The sulfur molecules bond to one another and release the hydrogens to create an interchain disulfide bond. This can also occur on 1 Cys molecule with 2 thiol groups in order to form an intrachain disulfide bond.

Question 64

Denaturation & Renaturation of Proteins

Answer 64

Proteins can be denatured using urea or excess heating which loosens it structural conformation by ‘breaking’ non-covalent interactions making the protein inactive or putting them in a reducing environment to break disulfide bonds. They can be renatured by removing urea or putting then into an oxidising environment which allows the protein to refold into its native configuration which is ‘memorised’ by the amino acid sequence. Not every protein can undergo this reformation.

Question 65

Secondary Structures of Proteins

Answer 65

A common motif of folding proteins is beta sheets. These are the rigid cores of proteins. The carbonyl oxygens on the polypeptide backbone in one beta strand form hydrogen bonds with the hydrogen on a nitrogen group of a second beta strand to form this beta sheet. These hydrogen bonds keep the beta strands together. The side chains stick out from the sheets and aren’t involved in holding sheets together. The beta sheets can be formed by beta strands in the same protein/polypeptide or between beta strands in different polypeptide chains.

Question 66

Parallel & Antiparallel Beta Sheets

Answer 66

In parallel examples the neighbouring beta strands run in the same orientation (both from N-C terminus or from C-N terminus). In antiparallel examples the neighbouring beta strands run in opposite directions (one going from N-C terminus and the other from C-N terminus). This also effects the structure of proteins.

Question 67

Alpha Helix

Answer 67

These are hydrogen bonds between the carbonyl oxygen atoms of a peptide bond and the amide hydrogen atom of the amino acid 4 residues away which stabilises the helical structure. The side chains stick outwards and aren’t involved in forming this secondary structure of proteins.

Question 68

Transmembrane Proteins

Answer 68

These types of proteins contain transmembrane domains. These domains are formed by alpha helices in which the hydrophobic (non-polar) side chains stick out into the hydrophobic lipid bilayers and the hydrophilic (polar) side chains face inward.

Question 69

Random Coils

Answer 69

Only a few proteins only have an alpha helix or beta sheet with most proteins have these unstructured units. These do not form regular secondary structure and aren’t characterised by any regular hydrogen bonding pattern. They are typically found at terminal ends (N or C terminus) and in loops which are found between regular secondary structure elements (alpha helixes and beta sheets. These are what allows for the flexibility of proteins in their secondary structure.

Question 70

Protein Domain

Answer 70

This is a motif is a region of a polypeptide that can fold independently into a compact, stable structure. The domain can be made up of 4 alpha helices, alpha helices and beta strands or from beta strands forming a beta sheet. There can be multiple of these motifs on a single protein with many different variations. The regions made with both alpha helices and beta strands are a single region and not 2 separate ones.

Question 71

Evolutionarily Related Proteins

Answer 71

A domain can occur in different proteins with the rest of the amino acid in the proteins being completely different. These domains are recognisable and listed in databases. These similar domains are found in proteins with a similar function in evolutionarily distant organisms. As an organism increases in complexity the number of domains will increase in their individual proteins (general rule not exactly true).

Question 72

Orders of Protein Structure

Answer 72

Primary Sequence: the amino acid sequence.
Secondary Structure: the local folded structures (alpha helix, beta sheet and random coil).
Tertiary Structure: the full 3D conformation of all alpha helices, beta sheets, random coils, loops of a polypeptide chain.
Quaternary Structure: the 3D relationship of polypeptides in a protein made up of more than one protein. Each of the proteins is referred to as a subunit (this refers to multi-subunit proteins only).

Question 73

Quaternary Structure

Answer 73

This has homodimers made up of 2 identical protein subunits and heterodimers made up of 2 distinct protein subunits. There is also trimers and tetramers etc. for proteins with even more protein subunits.

Question 74

Protein Function Examples

Answer 74

These are the agents of biological function and the expression of genetic information. Enzymes are used to catalyse chemical reactions. Structural proteins provide mechanical support to cells and tissues giving the cells shape and moving objects around within cells. Transport proteins which sit typically in a membrane and are used to transport metabolites around into or out of cells/organelles. Storage proteins are seen in seeds which will degrade in order to provide the necessary amino acids for a particular function e.g. germination of a seed. Signal proteins which transmit or recognise signals e.g. hormones. Special-purpose proteins these are unique, varied proteins which have many function e.g. proteins in plant cells which make animals sick when eaten.

Question 75

Protein Characteristics

Answer 75

The shape, flexibility and function is determined by the amino acid sequence. These are mostly globular. Chaperonins are proteins which help other proteins to open and close in the correct manner with certain proteins not requiring a chaperonin and being able to do this independently due to the amino acid sequence. A single chaperonin can effect multiple other proteins creating a protein megacomplex once attached. This also shows that proteins are very flexible which is required to undergo their many functions e.g. moving substances in/out of a cell.

Question 76

Ligands

Answer 76

These are other molecules that almost all proteins will bind to. These can be ions (Na+), small molecules (ATP) or macromolecules (other proteins or DNA). This binding is done at a specific binding site due to the complementary shape of both structures (side chain complementarity). This binding is non-covalent. The binding causes the protein to fold to provide a close fit with non-fitting ones falling out of the binding site because the sum of non-covalent interaction is too weak to fit it to the protein. Correct version will form many non-covalent interactions and create a tight fit.

Question 77

Antibodies

Answer 77

These are produced by the immune system against foreign molecules or antigens (ligands). 2 heavy and 2 light chains which both contain variable (different to other proteins) and constant (similar in all proteins) domains. These bind to antigens with the antigen binding site leading to the destruction of the antigen. Variable domains have variable loops to change the length and amino acid sequence -> this determines the specificity of this structure based on the antigen -> potentially billions of these being produced by an immune system.

Question 78

Structural Proteins

Answer 78

Keratin protein has alpha helical regions. These monomers assemble into dimers which is done by winding up both monomers to create a coil. These coiled dimers form staggered (head to tail arrangement of the dimers where there isn’t a complete overlap of the dimers) tetramers. 8 tetramers of keratin packed together form intermediate filaments. Keratin intermediate filaments are part of the cytoskeleton of epithelial cells that forms a scaffold from the nucleus to the edges of the cell. The packing together of the tetramers provides a solid support.

Question 79

Mutated Structural Proteins

Answer 79

There are several genetic disorders that are characterised by blistering of the skin due to mutation in the keratin gene. This is also seen in mice where a clumping and disruption of the keratin network in basal cells results in a blistering of the skin. This is because the keratin filaments don’t interact well and don’t hold the cell together creating holes (blisters).

Question 80

Enzymes

Answer 80

Typically the names of these proteins have the suffix ‘ase’ however there are some exceptions. The name usually indicates the substrate and nature of the catalysed reaction e.g. DNA polymerase shows that DNA is the substrate and it works to polymerise (make a polymer) that DNA.

Question 81

Protein Families

Answer 81

Most proteins have belong to large organisational groups e.g. kinase which has many different subgroups. The relationships between proteins is determined by comparing similarities and differences in amino acid sequences with more similar sequence having a closer relations (being closer in the evolutionary tree) and also considers function. This shows the trend of a common ancestor for which these evolutionary tree can be made for most proteins.

Question 82

Enzyme Functions

Answer 82

These are proteins (some can be RNAs) which catalyse chemical reactions which is done to speed up the reactions by reducing the activation energy required for that reaction to occur. These aren’t used up in the reaction and will be available to bind to new substrates and catalyse the same reaction repeatedly. Theoretically they can be present forever however due to the degradation of proteins this isn’t practically true. An example is carbonic anhydrase which speeds up the transfer of CO2 from cells to the blood which makes the reaction 107 times faster than without it present. These proteins are also substrate specific e.g. peptidase will only target specific peptide bonds in the target protein.

Question 83

Enzyme Process

Answer 83

The enzyme (E) binds to a specific ligand/substrate (S) forming an enzyme substrate complex (ES). The E catalyses a change e.g. cleavage of the substrate forming an E product complex (EP). The EP rapidly dissociates releasing the product (P) from E. The binding in this reaction is done through non-covalent interactions in the active sites. The organisation of atoms in the active site is optimised for catalysis. These enzymes speed up chemical reactions without being altered so can be used again.

Question 84

Enzyme Binding

Answer 84

This can occur in many different ways however these 3 are the major ways.
1. The enzyme binds to 2 substrate molecules and orients precisely to encourage a reaction to occur between them.
2. The binding of substrate to enzyme rearranges electrons in the substrate creating partial negative and positive charges that favour a reaction.
3. An enzyme strains the bond substrate molecule forcing it toward a transition state to favour a reaction e.g. bending or breaking the substrate.

Question 85

Enzyme Regulation

Answer 85

This can occur in many different ways including at the gene expression level, compartmentalising the enzymes, regulating enzyme degradation, by binding other molecules or by phosphorylation. The products of one enzyme catalysed reaction may be the substrates for another creating a metabolic pathway. A positive feedback will continue to create a particular product through enzymatic catalysing however an inhibitory feedback will create a build up a different protein meaning a different reaction may occur until the product of that becomes to highly concentrated leading to a different reaction being catalysed again.

Question 86

Enzyme Regulation (Binding Other Molecules)

Answer 86

Positive regulation may occur by a positive effector molecule ‘X’ which binds to a regulatory site of the protein which then increases the number of active enzyme molecules by also binding ‘X’ different to the substrate. Negative regulation can also occur where a product made late in a pathway acts as a negative effector molecule by inhibiting an enzyme catalysing a reaction early in the pathway by also binding to a regulatory site on the protein with feedback inhibition causing a common regulatory mechanism e.g. B -> X -> Y -> Z and Z stops catalysing the reaction of B -> X.

Question 87

Enzyme Regulation (Allosteric Regulation)

Answer 87

Enzymes regulated by effector molecules are allosteric proteins. This is most proteins which have 2 or more different conformations where activity is regulated by switching conformations. Regulatory and active sites of proteins can ‘communicate’. Binding of a positive regulator (allosteric activator) changes the shape of the active site so that the substrate can bind better. Binding of a negative regulator (allosteric inhibitor) changes the shape of the active site so the substrate binds less well or not at all.

Question 88

Enzyme Regulation (Phosphorylation)

Answer 88

This reaction occurs on amino acids which adds or removes a phosphate group. This process requires the amino acid has a hydroxyl group (serine, threonine or tyrosine) and phosphate from ATP. Protein phosphatase removes phosphate groups from the amino acid which reverses the reaction whereas protein kinase adds a phosphate group to the amino acid. This process can activate some proteins and inactivate others. The protein kinases and protein phosphatases involved in this process are also enzymes.

Question 89

Motor Proteins

Answer 89

Some of the major examples of these are kinesin and dynein move along microtubules (part of cytoskeleton). They move cytoplasmic components (cargo) along microtubules in opposite directions. ATP hydrolysis occurs at the head regions of these proteins whereas the cargo is bound at the tail regions of the proteins. The binding and hydrolysis of ATP is coupled with the movement of the proteins. The ATP binds causing conformational changes in the protein and the hydrolysis will then occur causing another conformational change causing the protein to move through the cell or along other proteins. This is a step by step process where each head of the protein is moved at a time from the hydrolysis and binding of an ATP molecule.