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biomolecules and enzymes Flashcards

1
Q

What are the 3 main features of monosaccharides?

A
  1. Have a free carbonyl** (C=O) group hence are all reducing sugars**
  2. Small size** and have multiple -OH* groups and can form multiple H bond with water
    ==> readily soluble in water** to be easily transported in animal and plant transport systems
  3. Ring structures exhibit alpha/beta-isomerism**
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2
Q

how are disaccharides formed?

how are disaccharides split?

A

made up of 2 monosaccharides joined by a glycosidic bond** by a condensation** reaction involving the loss of a water*** molecule

can be split by hydrolysis reaction** to break the glycosidic bond** via the addition of a water **molecule

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3
Q

what monosaccharides are sucrose, maltose, and lactose made up of ? (they are disaccharides)

A 
maltose = glucose + glucose
lactose = glucose + galactose
sucrose= glucose + fructose

they are all reducing agents except sucrose

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4
Q

structure of starch (plant storage polysaccharide) - amylose and amylopectin

  • monomers
  • glycosidic bond
  • orientation of monomers
  • structure of starch
  • bond between molecules
A

monomers: starch is made up of alpha glucose monomers

amylose
: made up of alpha glucose monomers linked by alpha(1-4) glycosidic bond
: shape: amylose is a helical molecule

amylopectin
: made up of alpha glucose monomers linked by alpha(1-4)g.b within a branch and linked by alpha(1-6)gb at branch points

: shape: amylopectin is helical and branched

: all alpha monomers in the chain have the same orientation
: no interchain hydrogen bonding

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5
Q

structure of glycogen (storage molecule)

A

monomers; made up of alpha glucose monomers

gb: alpha gluc monomers linked by alpha(1,4) gb within a branch and alpha(1,6)gb at branch points

orientation of monomers: all glucose monomers have the same orientation

shape: helical coil which is extensively branched

bonds between molecules( presence of inter chain H bond) : no interchain h bonding

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6
Q

structure of cellulose (structural molecule)

  • monomers
  • gb
  • orientation of monomers
  • shape of each molecule
  • bonds between molecules
A

monomers: made up of beta-glucose** monomers

Gb: beta glucose monoemrs linked by beta (1,4) gb**

orientation of monomers; alternate** beta-glucose monomers are inverted 180degree** wrt to each other

shape of each molecule: long, straight **chain

bonds between molecules: OH groups** projecting outwards in both directions of the molecule for interchain hydrogen bonding* between parallel chains –> forming microfibrils*

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7
Q

how structures of glycogen and starch good storage molecules?

(2 main points)

A
  1. many alpha glucose monomers are coiled into helices
  • INSOLUBLE IN WATER:
    most OH groups involved in intramolecular hydrogen bond within the helix hence few OH groups available for hydrogen bonding with water –> insoluble in water and the water potential of cells are unaffected by presence of glycogen and starch
    OR
  • LARGE ; Insoluble in water; osmotically inactive
  • LARGE yet COMPACT, ENERGY STORE
    —> ( structure ) alpha glucose monomers are linked by alpha(1,4) gb **which gives rise to a helical* molecules of amylose
    –> helical molecule so more glucose* units per unit volume, this compact* structure for storage*
  1. amylopectin and glycogen are branched
    - have many branched ends for hydrolytic enzymes to work on –> for many glucose***** molecules to be released at the same time; more ATP generated by respiration per unit time
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8
Q

how the structure of cellulose makes it a good structural molecule?

(4 main points)

specific STRUCTURE: FUNCTION

  • microfibrils: T_ S_
  • involved in interchain H bond : __
  • meshwork of microfibrils (porous structure) : F__ P___ + distribute stress… : S___ and R____
A
  1. Adjacent glucose units are inverted 180 with respect to each other and hence form a long, linear, unbranched molecule with free OH groups projecting out in both directions which can hydrogen bond with OH groups of other cellulose molecules lying parallel to it and form microfibrils.Hencemicrofibrils have high tensile strength.(Property of cellulose)
    - -> HIGH TENSILE STRENGTH
  2. cellulose molecules are involved in interchain hydrogen bonding hence few OH groups except for the surface of the microfibril to hydrogen bond with water
    - -> INSOLUBLE IN WATER; does not affect the water potential of cells
  3. The meshwork of microfibrils that form the cell wall (A) have a porous structure and hence the cell wall is freely permeable to water and solutes and allows movement of substances across the cell wall.

(B) Are STRONG AND RIGID and distributes stress in all directions to prevent the plant cells from bursting due to osmotic stress.

4.Cellulases that hydrolyse cellulose are found in very few organisms. Thus cellulose cannot be hydrolysed and used as a respiratory substrate and is a good structural molecule.

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9
Q

Formation of triglycerides

A

Three non-polar, hydrophobic**, hydrocarbon **chains are joined to a glycerol backbone via the formation of 3 ester linkages.

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10
Q

How are ester linkages formed?

A

between a hydroxyl group (-OH) and a carboxyl group (-COOH) via a condensation reaction. One water molecule is removed for each fatty acid joined to the glycerol.

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11
Q

What is the structure of glycerol and fatty acids?

A

Structure of glycerol: 3 Polar* hydrophilic OH groups which are able to hydrogen bond **with water : soluble in water

Structure of fatty acids: when ionized in water, has charged COO-group which can interact with water**: soluble in water

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12
Q

What are the properties of triglycerides

A

triglycerides are non-polar and hydrophobic; do not form hydrogen bonding with water molecules

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13
Q

What are the functions of triglycerides

name at least 3

A
  1. MAJOR FUNCTION - compact energy store
    : (structure 1 ) has a high proportion of C-H bonds and lower proportion of oxygen for an equivalent mass of carbohydrates from which energy in the form of ATP and metabolic water can be formed during oxidation

: Triglycerides release twice as much energy on oxidation compared to an equivalent mass of carbohydrates

: (structure 2:) long, non-polar hydrophobic hydrocarbon chains in triglyceride cannot form hydrogen bond with water making it insoluble; does not affect the water potential of cell

  1. Production of metabolic water - oxidation of triglycerides produce metabolic water which is particularly important for desert animals
    :: Lipids produce more metabolic water per unit mass when compared with an equivalent mass of carbohydrates

Other:

  1. Lipids are found beneath the layer of skin. They are poor conductors of heat and are able to provide thermal insulation to mammals especially those in cooler climates.
  2. Lipids are less dense than water and hence improve buoyancy in mammals, especially marine mammals like the whale.
  3. Lipids form a protective layer around delicate internal organs of mammals. Hence they act as shock absorbers and protect organs from mechanical damage.
  4. Lipids can function as a reservoir for storage of fat soluble vitamins E.g vitamins A, D and K.
  5. Lipids are insoluble in water hence osmotically inactive or without affecting water potential of mammalian cells.
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14
Q

What is the structure of phospholipid?

A

Each phospholipid consists of

  1. 2 long non-polar, hydrophobic hydrocarbon tails
  2. joined to a glycerol backbone via ester linkages
  3. with the third hydroxyl group of the glycerol backbone joined to a negatively charged(not polar!) phosphate group
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15
Q

What are the major components of the phospholipid bilayer of the cell membrane? What are the functions?

A
  1. the non-polar hydrophobic hydrocarbon tails face inwards, away from the water
  2. the hydrophilic, negatively charged phosphate group face outwards and interacts with the aqueous environment

–> gives rise to a phospholipid bilayer with a hydrophobic core that forms a selectively permeable membrane

(I) regulates movement of substances by acting as a barrier to ions and polar and large molecules

(ii) acting as a boundary between the intracellular & extracellular aqueous environment and
(iii) allows compartmentalization within a cell

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16
Q

How are polypeptides formed?

A

Amino acids are joined by a peptide bond via a condensation reaction with the removal of one water molecule.

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17
Q

How does a change in nucleotide sequence lead to a change in the function of the protein?

A

The nucleotide sequence in DNA determines amino acid sequence in polypeptide which determines types and locations of R groups which determines R group interactions which determines3D structure and function of protein.

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18
Q

What is the primary structure of proteins?

What is the bond that maintains the primary structure of proteins?

A

Primary structure: number and sequence of amino acids*** in a single polypeptide chain with amino acids linked by peptide bonds **
•Linear structure maintained by peptide bonds
•The sequence of amino acids (and theirR groups) in a polypeptide chain determines the type and location of chemical bonds/interactions*, and hence the 3D conformation and characteristics ** of a particular protein.

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19
Q

What is the secondary structure of proteins?

What is the bond that maintains the secondary structure of proteins?

A

Structure formed by regular coiling or pleating of a single polypeptide chain.
•Maintained by hydrogen bonds
(a) between C=O and N-H groups of the polypeptide backbone

note: R groups are not involved

20
Q

Describe the structure of alpha helix (secondary protein structure)

A

Made up of a single polypeptide chain which is wound into a coiled/spiral structure.
▪A hydrogen bond forms between the C=O group of one amino acid residue and theN-H group of another amino acid residue four amino acids away (!!) along the backbone of a single polypeptide
▪There are 3.6 amino acid residues in every turn of the helix

21
Q

Describe the structure of beta-pleated sheet

A

Two or more regions/segments of a single polypeptide chain lying side by side are linked together by hydrogen bonds**
▪A hydrogen bond Forms between the C=O group of an amino acid residue and a N-H group of another amino acid residue on an adjacent segments along the backbone of a single polypeptide
▪Chains may run parallel(same direction) or anti-parallel(opposite directions)
▪Forms flat sheet which becomes folded

22
Q

What is the tertiary structure of proteins?

What are the bonds involved? (formed between?)

A

Structure formed by further extensive folding** and bending* of a single polypeptide chain, ** giving rise to the specific 3D conformation** of a protein
•Maintained by all 4 types of interactions
hydrogen bonds,
ionic bonds,
hydrophobic interactions*
and disulfide bonds*

  • formed between R groups** of amino acid residues within a polypeptide**
23
Q

What is the quaternary structure of proteins?

A

Refers to the association of two or more polypeptide chains** into one functional protein molecule
•Maintained by all 4 types of interactions
hydrogen bonds, ionic bonds, hydrophobic interactions and disulfidebonds
- formed between R groups ** of amino acid residues of different polypeptides

24
Q

What are the 4 types of bonds formed between proteins?

A
  1. Disulfide bond - Disulfide bridge between sulfhydryl groups of cysteine residues
  2. Hydrogen bond - between polar R groups
  3. Ionic bond - between oppositely charged R groups
  4. Hydrophobic interaction: formed between non-polar hydrophobic R groups
25
Q

What are the 3 structure:function of haemoglobin?

A
  1. (structure) Haemoglobin has a quaternary structure** made up of 4 polypeptide subunits: 2 alpha- globin subunits* and 2 beta-globin subunits*. Each subunit is made of globin polypeptide and a prosthetic (non-protein) component called haem group. Each haem group consists of a porphyrin ring and an iron ion (Fe2+)

(Function): Fe2+of haem group binds temporarily** to O2, so 1 Hb molecule can carry up to 4 O2, at a time forming oxyhaemoglobin

  1. (structure) Each subunit is arranged so that most of its hydrophilic ** amino acid side chains are on external surface** while its hydrophobic amino acid side chains are buried in interior
    (function) : haemoglobin is soluble in an aqueous environment **     and can be transported in the blood while carryingO2from lungs to tissues vice versa
  2. (structure) The 4 subunits held together by intermolecular interactions formed between R groups* (hydrogen bonds, ionic bonds and hydrophobic interactions). No disulfide bridges.
    (function) As a result binding of one oxygen molecule to one haemoglobin subunit induces a conformational change ** in remaining 3 subunits so that their affinity for oxygen increases    * . This is known as the**cooperative binding of oxygen.
26
Q

Describe the structure of a collagen molecule [4]

A
  1. A collagen molecule is made up of 3 polypeptide chains** tightly wound together to form a tropocollagen** molecule which is a triple helix
  2. Numerous H bonds** formed between the adjacent polypeptide chains
  3. between the NH group and the CO groups of amino acids as well as between OH groups of hydroxy proline residues
  4. Every 3rd amino acid in the polypeptide is a glycine** residue which is found on the inside of each polypeptide
27
Q

What are the 5 structure:function of collagen (fibrous protein)?

A
  1. (structure) A tropocollagen molecule consists of three helical polypeptide chains(loose helices, not α-helices) wound around each other like a rope. (has quaternary but no tertiary structure)
  2. (structure) Each chain contains about 1000 amino acids and contain a repeating sequence, usually a repeating tripeptide unit: glycine-X-Y,where X is usually proline,Y is usually hydroxyproline. The tropocollagen molecule can form a tight,compact coil as almost every third amino acid in each polypeptide chain is a glycine,the smallest amino acid. This allows it to fit into the restricted space in the center of the triple helix.
    (function) : Bulky and relatively inflexible proline and hydroxyproline residues confer rigidity on the molecule
  3. (structure): Extensive hydrogen bonds form between amino acid residues of adjacent polypeptides, hence interaction with water molecules is limited.
    (function) : Increases tensile strength (ability to resist snapping due to stretching) and makes the molecule insoluble in water (extensive H bond already formed between residues of polypeptides hence interaction with water molecules are limited)
  4. (structure) Adjacent tropocollagen molecules are arranged in a staggered manner
    (function) : Staggered/overlapping arrangement minimizes points of weaknesses along fibrils
  5. (S) Covalent cross-links between lysine residues at C and N ends of adjacent tropocollagen molecules results in the formation of fibrils.

(F) Greatly increases tensile strength

6.(S) Bundles of fibrils unite to form long collagen fibres.

(F) Large size of collagen makes it insoluble, an important property for a structural molecule

28
Q

What is denaturation of protein?
(occurs as a result of disruption of interactions)

Which level of structures do denaturation affect

What are the factors leading to denaturation?

A

Loss of conformation of a protein molecule causes the protein to lose its function

  • secondary structure and above
  • High temperatures: increase KE and intramolecular vibrations –> breaks interactions such as H bonds
  • Extreme pH –> affect ionisation of R-groups of charged amino acids and hence ionic bonds
29
Q

Differences in solubility of fibrous proteins and globular proteins

A

1) solubility in water

Globular proteins : soluble in water—> as polar R groups ** can form hydrogen bonds w water molecules in aqueous environment **

Fibrous proteins are insoluble in water —> large molecule* and extensive hydrogen bonds* have already formed between residues* in different polypeptides

2) shape
Globular: made up of polypeptide chains folder into roughly spherical* shape

Fibrous: made up of long polypeptide chains forming long* straight* fibres*

3) amino acid sequence

Globular: more variety of amino acids** are used to construct the protein** (I.e consists of non-repetitive amino acid sequence)

Spherical: less variety of amino acids** used to construct the protein (i.e. consists of repetitive regular sequence ** of amino acids (e.g. tripeptide : glycine-X-Y repeats in collagen) )

4) Length of polypeptide

Globular: length of polypeptide is always identical between 2 samples of the same protein, or else protein may not be functional)

Fibrous: length of polypeptide may vary slightly* between the 2 samples of the same protein, yet protein is still functional

5) function

Globular: protein with metabolic role (e.g enzyme)

Fibrous : structural role

30
Q

Definition of an enzyme

A

KW:

  • biological catalysts
  • increase rate of reaction
  • chemically unaltered (at the end of the reaction)
  • can be reused
  • effective in small amounts
31
Q

Explain the induced fit hypothesis

A

—> the active site** of the enzyme is not exactly complementary in shape and charge** to the substrate
→binding of the substrate to the enzyme active site induces a conformational change in the active site of the enzyme such that it now provides a more precise fit for the substrate
→thus enzyme can perform its catalytic function more effectively; allows for the FORMATION OF ENZYME SUBSTRATE COMPLEX

31
Q

What are the 3 different types of enzyme cofactors?

some enzymes require additional non-protein substances for catalytic activity

A
  1. Inorganic ions
    e. g. Ca2+, Zn2+
    - attachment of the ion with the main enzyme changes the shape of the enzyme so as to allow the ES-complez to form more easily
  2. Coenzymes
    e. g NAD
    - organic molecules required by certain enzymes to carry out catalysis
    - bind to active site of the enzyme and participate in catalysis but are not considered substrates of the reactions
    - function as intermediate carriers of electrons or specific atoms that are transferred in the overall reaction
  3. Prosthetic Group
    (e. g. haem group of cytochrome oxidase in ETC in inner mitochondrial membrane)
    - permanently bound to the enzyme
    - transfers atoms/ chemical groups between active site of enzyme in another substance
32
Q

Using the induced fit model, explain how hexokinase binds to glucose [2]

A
  1. Based on induced fit hypothesis, Glucose binding to active site causes a change in shape* in active site**
  2. So that active site* is a more precise fit for substrate* for effective catalysis
  3. Enzyme substrate complex held together by weak interactions e.g. hydrogen, ionic bonds**
33
Q

What are the different types of amino acid residues making up the enzyme?

(ALSO: how does the structure of the active site of the enzyme ensure correct binding with substrate and subsequent catalysis? )

A
  1. Contact residues - found in the active site–help to position the substrate in the correct orientation via weak interactions such as hydrogen bonds, ionic bonds &hydrophobicinteractions.
  2. Catalytic residues - found in the active site–have specific R groups which act on bonds in the substrate and help to catalyse the conversion of substrate to product (e.g facilitates the cleavage of glycosidic bond)
  3. Structural residues - interact with each other to maintain the overall 3D conformation of the protein
34
Q

How do enzymes speed up metabolic reactions by lowering activation energy?
(what are the 5 effects?)

A
  1. Proximity effect - temporary binding of substrates in close proximity in the enzyme active site increases chances of a reaction (vs depending only on random collisions between reactants in the absence of enzymes)
  2. Strain effect - slight distortion of substrates as they bind to enzyme strains bonds in substrates that need to be broken for products to form
  3. Orientation effect - substrates are held by enzymes in their active sites in an the correct orientation such that the bonds in substrates will be exposed to chemical attack
  4. Microenvironment effect - enzymes create the appropriate microenvironment for a reaction to occur e.g. hydrophobic amino acids in enzymes create a water-free zone that allows non-polar reactants to react more easily
  5. Acid-base catalysis - R groups of acidic and basic amino acids in enzyme active site facilitate reaction between substrates

–> for greater proportion of substrate molecules with energy greater than or equal to activation energy, allowing rate of reaction to increase

35
Q

What is the temperature coefficient? Q10?

A

Q10= Rate of reaction at (x+10)oC/ Rate of reaction at xoC

36
Q

What is the effect of temperature on enzymes?

A

As temperature. increases,
➔Kinetic energy** of enzyme and substrate molecules increases➔Frequency of effective collisions ** between enzyme and substrate molecules increases➔Rate of enzyme-substrate complex** formation increases➔ Number of substrate molecules with sufficient energy** to overcome the activation energy barrier and form products increases and rate of reaction **increases.

When optimum temp exceeded,
➔kinetic energy** of enzyme and substrate molecules continue to increases➔intramolecular vibrations increases➔hydrogen bonds, ionic bonds and hydrophobic interactions ** that maintain the 3D conformation of the enzyme are disrupted( Covalentdisulfidebonds are harder to break and thus can withstand higher temperature. Hence enzymes with higher optimum temperature tend to have a larger proportion of disulfide bridges or more intramolecular interactions.)➔specific conformation of active site is lost➔substrate no longer complementary* to the shape and charge of active site** and cannot bind to it➔rate of enzyme-substrate complex** formation decreases and rate of reaction decreases.

37
Q

What is the effect of pH on enzymes?

A

At optimum pH,➔conformation of enzyme active site is most ideal for substrate binding and rate of reaction is highest

When optimum pH exceeded,
➔excess H+or OH-ions affects the ionisation of the R-groups** of the amino acids residues as excess H+results in –COO-groups becoming -COOH and excess -OH-results in -NH3+becoming –NH2
➔thus ionic bonds * and hydrogen bonds** that maintain the conformation** of the enzyme active site ** is disrupted➔thus the interaction between substrate and catalytic residues in the active site of enzyme is disrupted➔rate of enzyme-substrate complex** formation decreases➔ rate of reaction/product formation decreases
(Note: if the change in pH affects the charges of the R groups of the
(1)catalytic residues in the active site, the catalytic activity of enzyme may be lost
(2)contact residues in the active site, this may affect the temporary binding between the enzyme and substrate and thus no enzyme-substrate complex forms.
(3)structural residues, the tertiary structure of the protein and its active site can be affected and this would denature the enzyme.)

38
Q

Based on the graph, describe the effect of increase in [enzyme]

  • at linear portion
  • at the curved portion
  • at the plateau
A

Initially, when [enzyme] is low, as [enzyme] increases,➔frequency of effective collisions** between enzyme and substrate molecules increases ➔rate of enzyme-substrate complex *formation increases and rate of reaction increases

At linear ** portion of graph, [enzyme] is limiting➔increasing[enzyme] will result in a proportional ** increase in the rate of reaction

At curved portion of graph, rate of reaction is increasing gradually/at a decreasing rate.
[Enzyme] is not the only limiting factor. Some other factor is also limiting

At the plateau, [enzyme] is no longer the limiting factor** (other factors are limiting)➔increasing [enzyme] will not affect the rate of reaction

39
Q

Based on the graph, describe the effect of increase [substrate]

A

Initially, when [substrate] is low, as [substrate] increases,➔ frequency of effective collisions ** between enzyme and substrate molecules increases➔ rate of enzyme-substrate complex ** formation increases and rate of reaction increases as active sites** of enzymes are readily available ** and substrate concentration is limiting**

Beyond a certain [substrate],
➔all active sites of enzymes are saturated* with substrate at any one point in time
➔[substrate] is no longer limiting ** and enzyme concentrationis limiting , the rate of reaction will remain constant(graph plateaus)and reaches maximum velocity (Vmax).

40
Q 
Michaelis Constant (Km) 
- relevant to effect of [substrate] on rate of reaction
A

[substrate] at which reaction proceeds at half its max. rate
- it is the measure of affinity of the enzyme for its substrate
➔low Km –high affinity between enzyme & substrate
➔high Km –low affinity between enzyme & substrate

41
Q

Explain how competitive inhibition occurs

Effect of increasing [substrate] on the inhibition

Can the effect of increasing [substrate] overcome inhibition?

A

Inhibitor competes with substrate** for active site **of enzyme as they are both structurally similar➔This reduces the availability ** of active sites for substrate binding➔rate of reaction decreases

At high [substrate], the effect of the inhibitor is negligible since the higher proportion of substrate ** molecules compared to the inhibitor can effectively outcompete ** the inhibitor* molecules for the active site**
➔Thus the effects of inhibition ** can be overcome** and Vmax can be reached * at high[substrate]

Vmax same, Km increases

43
Q

Explain how non-competitive inhibition occurs

Effect of increasing [substrate] on the inhibition

Can the effect of increasing [substrate] overcome inhibition?

A

Inhibitor binds to a site other than the active site**
➔This results in a conformational change** in the enzyme active site**
➔Thus substrate can no longer bind to active site
➔rate of reaction decreases
➔Hence the inhibitor effectively decreases the available [enzyme] as it forms an inactive** enzyme-inhibitor complex **
➔Hence the effects of the inhibition cannot be overcome** by increasing [substrate]

Vmax decreases, Km remains the same

44
Q

Explain how allosteric inhibition works

(S): inhibitor/activator complementary to the allosteric site

: multimeric enzyme with multiple active sites and allosteric sites
: enzyme freely oscillates between the active form and the inactive form

Effect of increasing [substrate] on the inhibition

A

Inhibitor/Activator binds to allosteric site of the enzyme. This results in conformational change in enzyme.

•Binding of inhibitor stabilise enzymes in an inactive state
➔shifts curve to the right

•Binding of activator stabilises enzyme in an active state
➔shifts curve to the left

increase [substrate],
Substrate binding stabilizes the enzyme in the active conformation and opposes the effect of the inhibitor.This allows Vmax to be reached at high substrate concentration.

•Binding of substrate in allosteric enzymes exhibit cooperativity.
Binding of a substrate to the first subunit, changes the conformation of the other subunits such that it is easier to accept subsequent substrates.

(GRAPH) Hence the rate against substrate concentration plot is sigmoidal.

45
Q

End-product inhibition

A

→Regulation of a metabolic pathway by the end-product of the pathway
→End-product can function as an allosteric inhibitor to an enzyme present early in the pathway by binding to its allosteric site and prevent further synthesis of the product.

46
Q

What are polysaccharides

A

Made up of many monosaccharides joined by glycosidic bonds** formed between them condensation reactions** which involve the loss of water molecules**

47
Q

What is the lock and key hypothesis

A

The enzyme’s (lock) active site has a conformation* which is complementary in shape and charge** to a specific substrate** ( the key)

allowing substrate to bind, hence forming an enzyme-substrate complex ***

Products formed are no longer fit** for the active site and will hence leave the active site, making it available for another substrate to bind

JC bio (16 decks)

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