3.1 Chapter 1- Biological Molecules Flashcards

Question

What are the different types of glucose?

Answer 1

* Glucose has two structural isomers, α-glucose and β-glucose.

Answer 2

* Answer on revision card. * Remember alpha glucose the H is above the ring, in beta glucose the H is below the ring. * Important to be able to identify them and also remember which forms cellulose vs. which forms starch/ glycogen.

Answer 3

1. Dissolve the sample 2. Add an equal volume of Benedicts reagent 3. Heat to over 70°C 4. If present, the solution will turn from blue to brick red All monosaccharides are reducing sugars. Test works by reducing Cu²⁺ ions to Cu⁺ turning them from blue to red. High concentration means more red substrate.

Answer 4

* You could use chromatography, with known standard solutions, to separate a mixture of monosaccharides and identify their components.

Answer 5

* Chromatography- used to separate a mixture into individual components. * Chromatography- relies on differences in solubility of different solutes within a mixture. * Has two phases- the mobile phase and the stationary phase. * Components in the mixture separate as the mobile phase travels over the stationary phase. * Differences in the solubility of each component in the mobile phase affects how far each component can travel. Higher solubility= travel further- spend more time in the mobile phase- carried further up the stationary phase. * In paper chromatography, the mobile phase is a solvent e.g. water or ethanol, and the stationary phase is chromatography paper.

Answer 6

* Stain the monosaccharides first. * Add a spot of stained monosaccharide on the origin line drawn at the bottom of the chromatography paper. * Spots of known standard solutions of different monosaccharides are placed on the line beside the sample. * The chromatography paper is suspended in a solvent. * As the solvent travels through the chromatography paper, different monosaccharides within the mixture separate at different distances from the line. * Molecules separate at different speeds- larger molecules travel slower than smaller ones- causes the mixture to separate into different spots or band son the chromatography paper. * The unknown monosaccharides can be identified by comparing and matching them with the chromatograms of the known standard solutions different monosaccharides. * If a spot from a monosaccharide sample has the same Rf value as a spot from the standard solution, the mixture contains that monosaccharide.

Answer 7

- Commonly used by cells as respiratory substrate - Form structural components in plasma membranes and cell walls

Answer 8

* Formed by the condensation reaction of two monosaccharides, forming a glycosidic bond and water. * Glycosidic bonds are formed between the **two OH molecules** on alpha glucose-** may need to label them. ** * Broken by hydrolysis requiring water.

Answer 9

* Reducing or non-reducing sugars. * Reducing sugars- can donate electrons to other molecules reducing them- is oxidised. Include monosaccharides- fructose, galactose, glucose. * Non-reducing sugars- can’t be oxidised to give electrons that reduce other molecules. Include disaccharides such as sucrose and other carbohydrate molecules.

Answer 10

* Maltose-disaccharide formed by the condensation of **2 alpha glucose molecules**- forms **α1-4 glycosidic bond.** * Sucrose- disaccharide formed by the condensation of an alpha glucose and fructose molecule. * Lactose- alpha **glucose and galactose**. * **Each have glycosidic bonds. **

Answer 11

**2α glucose —> maltose** + water 2C₆H₁₂O₆ —> C₁₂H₂₂O₁₁ + H₂O

Answer 12

Answer on revision card Remember to label **α1-4 glycosidic bond**

Answer 13

* Using Benedict’s solution. * This is a qualitative test but a semi-quantitative value can be produced using the precipitate and the colour change as an approximate indication of how much reducing sugar is present knowing the exact quantity. * Should be able to use, and interpret the results of, qualitative tests for reducing sugars, non-reducing sugars and starch.

Answer 14

1. Perform the test for reducing sugars make sure it’s negative 2. Take new sample, dissolve and **add** an equal volume of HCl/ **acid** 3. Boil/ **heat** to hydrolyse the glycosidic bonds in the disaccharide 4. Add sodium hydrogen carbonate to **neutralise**. Test the solution is alkaline using indicator paper. 4. **Add Benedict’s reagent** and **heat** over 70°. If measuring quantitively, it is important to ensure there is an excess of Benedict’s solution so there is more than enough copper II sulphate present to react with any sugar. 5. They reducing sugar is present. The solution turns from blue to brick red. **Red precipitate/ colour**.

Answer 15

Semi quantitative

Answer 16

* * The intensity of the colour change relates to the concentration of reducing sugar present in the sample. Spectrum goes from blue to brick red. * High concentration means more red substrate * Measured by **filtering, drying and finding the weight of the precipitate**. * **In disaccharides, there will be more sugar present after hydrolysis- more precipitate so darker red-** another way to tell the presence of a disaccharide in a solution. * Can be measured using a colorimeter. * Could produce a dilution series of glucose solution and use colorimetric techniques to produce a calibration curve with which to identify the concentration of glucose in an unknown solution.

Answer 17

* Used to measure the colour change of the Benedict’s reagent test and therefore the concentration of reducing sugars. Measures colour change using the absorption of transmission of light. * Makes the test more **quantitative** to make the investigation more repeatable- **standardises the method and makes it less subjective and more accurate/precise** than measuring by eye **as colour change is subjective.** * In general measures the percentage absorption or transmission of a particular wavelength of light emitted by a solution

Answer 18

Reduce blue Cu²⁺ to red Cu⁺ in the copper II sulphate solution forming a red precipitate.

Answer 19

1. The Benedict’s solution is filtered to remove the red precipitate, leaving just the blue solution. The more glucose means the solution is lighter blue as the solution is more reduced 2. The cuvette is filled 3/4 full of blue liquid and a separate cuvette is filled 3/4 full of distilled water 3. The colorimeter is set to red and the water is placed inside to calibrate it to 0. The solution is then added and the percentage absorption is measued. 4. The red light is absorbed by the blue as this is a complimentary colour, the less blue that means the more red light transmitted and less absorbed. This means more red precipitate was produced, and there was a great quantity of reducing sugar.

Answer 20

* Semi-quantitative test can be done by using standard solutions with known concentrations of a reducing sugar such as glucose. * Solutions should be set up using a serial dilution of an existing solution. * Each solution is treated the same way- add same volume of Benedict’s and heat at the same temperature at the same time to allow colour changes to occur. Important to ensure there is an excess of Benedict’s solution. * Any colour change observed can be attributed to the concentration of reducing sugar present in the solution. * You can find the concentration of a sample of unknown concentration of reducing sugars by comparing it with the stock solution colours to estimate the concentration of reducing sugars present.

Answer 21

The results of the unknown solution will be used to plot a calibration curve at different glucose concentrations, which can then be used to find the glucose concentration of an unknown substance by comparing it.

Answer 22

* Created by taking a series of dilutions of a stock solution. * The concentration decreases by the same quantity between each test tube- either be doubling dilutions or a desired range. * Serial dilutions are compared to create a standard to compare unknown concentrations against. * See chapter 4 for more info on serial dilutions.

Answer 23

* Formed in the condensation reaction of many monosaccharide monomers or glucose units, forming water and glycosidic bonds. * Insoluble.

Answer 24

Hydrolyse into disaccharides or monosaccharides

Answer 25

* Polysaccharide made from the condensation of repeating units of **α glucose** joined together by glycosidic bonds in a condensation reaction * Made of amylose (unbranched and coiled helical structure due to hydrogen bonds- α 1-4 glycosidic bonds ) and amylopectin (branched- α 1-4 and α 1-6 glycosidic bonds).

Answer 26

Starch is **hydrolysed into maltose by amylase.** This enables the maltose to be **absorbed** and transported around the body and enables it to be further hydrolysed into glucose for respiration as starch cannot be absorbed and is egested.

Answer 27

* Storage of alpha glucose in plants, e.g. seeds. * **Polymer of α glucose- provides glucose for respiration.** * **Insoluble in water- doesn’t affect water potential or osmosis.** * **Large molecule- can't cross cell membrane and leave cell.** * **Compact- coiled and branched, helical** structure of amylose stores **lots o**f glucose **molecules in a small area.** * **Branched** endings- **more ends- fast enzyme action/ breakdown-** easily hydrolysed for rapid glucose release due to branched ends of amylopectin meaning enzymes can attach and work simultaneously.

Answer 28

* Add two drops of **iodine/ potassium iodide** and see if the colour turns from brown orange to black blue. * Useful to show if starch in a sample has been digested by enzymes.

Answer 29

* **Polysaccharide polymer** made of **α glucose monomers** joined together in a condensation reaction to form glycosidic bonds. * **α 1-4 and α 1-6 glycosidic bonds. ** * Shorter chained and more highly **branched** than starch. Similar structure to amylopectin. * Some of the chains in glycogen are **coiled**.

Answer 30

* Used for alpha glucose/carbohydrate storage in animals and bacteria. e.g. in the liver. **Hydrolyses into glucose which is used in respiration**. Small amount of glycogen because main storage molecule in animals is fat. - Insoluble- Doesn’t affect water potential or osmosis - Compact- stores lots of glucose in a small space - Highly branched- made of amylopectin, More ends to be acted on simultaneously by enzymes. More rapid as animals have a higher metabolic rate so quicker energy release.

Answer 31

- * Polysaccharide formed from the condensation of **β glucose monomers.** - **Position of hydrogen and hydroxyl groups on carbon atom 1 inverted** every other monomer - * **Only β 1-4 glycosidic bonds.** - **Straight unbranched chains**, parallel to each other - Joined together by hydrogen bond cross linkages to add strength- weak on their own but strong together - Group to form strong, microfibrils and fibres

Answer 32

- Forms plant cell walls - Provide structural support and rigidity to the plant cell - **Long straight chains linked together by many hydrogen bonds to form fibrils that provide strength** - The cell from bursting under osmosis and exerts an inward pressure to stop water influx - Makes stems and leaves turgid and rigid, creating a maximum surface area for photosynthesis - Permeable to water- allows water movement

Answer 33

* To include both items in each point of comparison- use whereas/ and terms.

Answer 34

Insoluble in water, soluble in organic solvents

Answer 35

* Cell (plasma) membranes- help form bilayer * Used in certain hormones and respiratory structures. * Energy source- 2x energy as glucose and produces water * Waterproofing- waxy cuticle and oil on the skin * Electrical and heat insulation * Protecting delicate organs

Answer 36

* Made of **one glycerol and three fatty acids** * Formed in a **condensation reaction** forming **ester bonds** and **three molecules of water**. Reversed by hydrolysis.

Answer 37

* A **fatty acid is made of a carboxyl (COOH) and an R group** * Variation is found in the R group of fatty acids * Saturated- no carbon to carbon double bonds- saturated by hydrogen * **Unsaturated- carbon to carbon double bonds**

Answer 38

The 3 fatty acid tails are hydrophobic.

Answer 39

* Vary in the R group of fatty acids. * Saturated- no carbon to carbon double bonds- saturated by hydrogen * **Unsaturated- carbon to carbon double bonds**

Answer 40

* Good source of energy- high ratio of energy storing carbon-hydrogen bonds to carbon. Produce two times energy of carbohydrates * Low mass to energy ratio- high energy in a small volume- reduces the mass animals have to carry. * Insoluble- large and non-polar and don't affect osmosis or water potential. * Important source of water- high ratio of carbon to oxygen, so release water when oxidised.

Answer 41

* Form Insoluble droplets aka micelles * See revision card for drawing

Answer 42

Answer on revision card.

Answer 43

* Similar to triglycerides, except one fatty acid is replaced with a phosphate. * Hydrophilic phosphate heads- attract water- interact with water but not fat- negatively charged. * Two hydrophobic fatty acid tails- repel water- orient away from water but mix with fat * Polar molecule- head as close to the water as possible. Tails far away.

Answer 44

* Polar molecules form a bilayer. * Phosphate heads face outwards, fatty acid tails face inwards. * Helps form the cell surface membrane * Forms a hydrophobic barrier- the centre blocks water soluble substances

Answer 45

* Provide flexibility * Transfer lipid soluble substances * The hydrophilic phosphate heads hold the phospholipids at the surface of the cell-surface membrane.

Answer 46

Glycolipids with carbohydrates

Answer 47

'Drawing on revision card Labels should include: Uncharged hydrophobic fatty acid 'tails' Negatively charged hydrophobic phosphate 'head'

Answer 48

1. Take a dry, grease free test tube 2. **Add** a sample and double the volume of **ethanol** 3. **Shake** to **dissolve** lipid in sample 4. **Then add** equal volume of **water** to ethanol and shake 5. If a **milky white emulsion** forms lipids are present Control- process repeated using water instead of sample- solution should be clear.

Answer 49

The monomers which proteins are made from through condensation reactions.

Answer 50

Evolution- There are only twenty different types of naturally occurring amino acids.

Answer 51

* **Amine (NH₂) group** at one end * **Carboxyl group (COOH)** at other end * Single hydrogen molecule and carbon at centre * **R group**/ R side chain- **where amino acids differ**/ vary

Answer 52

Answer on revision card Labels should include: Amine group Carboxyl group R side chain- only place amino acids differ

Answer 53

* Forms a peptide bond between the **amine and carboxyl groups** * Creates a dipeptide * **Releases water.** * Peptide bonds can be broken by hydrolysis using water.

Answer 54

Answer on revision card.

Answer 55

Amino acids form polypeptide chains that form proteins.

Answer 56

Proteins form many cell structures. They are important as enzymes, chemical messengers and components in the blood.

Answer 57

A protein shape determines its function. Structure and shapes have to be specialised e.g. enzymes

Answer 58

Polymerised in condensation reactions to form polypeptides and **peptide bonds** **between the amine** (NH₂) **group** and the **carboxyl group** (COOH). Each amino acid is orientated in the same direction and **in each chain there is a free amine group at one end and a free carboxyl group at the other**.

Answer 59

One or more polypeptide chains of amino acids. Often more than one.

Answer 60

One polypeptide chain

Answer 61

**The relative positions of amino acids and the bonds between them (hydrogen, disulfide bridges, ionic)**.

Answer 62

**The sequence of amino acids** in a polypeptide chain- specific order.

Answer 63

DNA and genes that code for the protein- in base sequences of triplets- every three bases codes for one amino acid.

Answer 64

- The sequence of amino acids determines the end shape and function of the protein. - A change in just one amino acid along sequence changes the shape of the whole protein and stops it from functioning. - The protein shape is specific to its function. - A change in the primary structure produces different **bonds in different places (ionic, hydrogen, disulfide bridges)** that produces a different teritary structure.

Answer 65

The **same amino acids can form different sequences- primary structures**.

Answer 66

**Hydrogen bonds form between amino acids**. Form **between NH**/ the hydrogen in the amine group **and** the **double bonded oxygen and carbon** in the carboxyl group. Forms a twisted 3D shape. e.g. **alpha helix/ beta pleated sheet.**

Answer 67

- More complex and specific 3D structure. - **Formed by** further twisting coiling and folding the polypeptide. **Bonds between amino acids** do this. - **Disulphide bridges** are strong and not easily broken, formed from two sulfurs. - **Ionic bonds** are formed from the carboxyl and amine group not in peptide bonds- easily broken by a pH change. - **Hydrogen bonds** are common and weak.

Answer 68

The tertiary structure.

Answer 69

Several (**more than one**) different linked **polypeptide chain** **formed by bonds between polypeptides (hydrogen, disulfide bridges, ionic bonds)**, the way they are assembled together and **bonds to** non-protein **(prosthetic) groups** associated with the molecules.

Answer 70

Although 3D structures important to have protein functions. The sequence of amino acids determines the 3D shape.

Answer 71

The biuret test: 1. Add a few drops of sodium hydroxide to make the solution alkaline. 2. Then **add** copper II sulphate (**biuret reagent**) 3. If proteins are present, the solution turns from blue to **purple**.

Answer 72

Enzymes are biological catalysts and globular proteins

Answer 73

Biological catalysts that increase the rate of reaction of metabolic reactions without undergoing permanent change and can therefore be reused.

Answer 74

* Catalyse metabolic reactions: * At a cellular level e.g. respiration and for the organism as a whole e.g. digestion. * Help produce structures, e.g. proteins and functions, e.g. respiration. * Functions can be intracellular (in the cell) or extracellular (outside the cell).

Answer 75

Enzymes are globular proteins with a specific 3D tertiary structure as a result of their primary structure (sequence of amino acids)

Answer 76

The functional region of an enzyme with a specific shape where substrates bind to. A small depression made of a few amino acids specific to the substrate.

Answer 77

Molecules on which enzymes act. Fits into the active site held by temporary bond between amino acids and the substrate to form enzyme substrate complexes.

Answer 78

The minimum amount of energy needed for a reaction to start. Often in the form of heat.

Answer 79

The activation energy and a successful collision.

Answer 80

1. Catalyse reactions 2. Form enzyme substrate compexes 3. **Lower the activation energy** 4. **Induced fit causes the active site of the enzyme to change shape** 5. **The enzyme substrate complex causes bonds to form/ break** by **putting tension on/ bending the bonds or weakening the bonds.** 6. Joining- being attached to the enzyme holds the substrates close together and reduces the repulsive forces. Breaking- the active site put strain on the bonds of the substrate, so they break more easily.

Answer 81

Enzymes lower the activation energy so reactions can happen at lower temperatures and this speeds up the rate of reaction. This enables metabolic processes to happen rapidly at the low body temperatures e.g. 37°C. Without this reactions would be too slow to sustain life.

Answer 82

Answer on revision card

Answer 83

Joining- overcoming repulsion forces between molecules by holding them close together. Breaking- puts strain on the bonds of the substrate to break it apart.

Answer 84

- Accepted theories change as new evidence comes to light. - For example, the lock and key model, where the enzyme and substrate have rigid complementary shapes has been replaced by the induced fit model where the enzymes shape changes slightly.

Answer 85

The enzyme is flexible and mould around the substrate. This only occurs with the specific substrate the enzyme is complementary to otherwise, the enzyme maintains its usual shape. As the enzyme changes shape, it puts strain on the substrate and reduces the activation energy.

Answer 86

1. **Substrate binds to the active site and enzyme-substrate complexes form** 2. **Active site changes shape (slightly) so it is complementary to substrate** 3. **This breaks/forms bonds in the substrate** 4. **This reduces activation energy**

Answer 87

In the presence of the enzymes complimentary substrate.

Answer 88

Joining- being attached to the enzyme holds the substrates close together and reduces the repulsive forces. Breaking- the active site put strain on the bonds of the substrate, so they break more easily.

Answer 89

The tertiary structure of the active site.

Answer 90

Highly specific. Usually only catalyse one reaction and have only one complimentary substrate that will fit into the active site.

Answer 91

The tertiary structure, which is determined by the primary structure (sequence of amino acids). Different for different types of enzyme.

Answer 92

Enzyme substrate complexes do not form and the reaction isn’t catalysed.

Answer 93

- The shape of the active site changes. - The substrate won’t fit as the active site is no longer complimentary. - Enzyme substrate complex are not formed. - The enzyme can’t carry out its function and catalyse the reaction. Caused by: - pH or temperature changes - Changes in the primary structure due to mutations causing a change in the tertiary structure.

Answer 94

Change the shape of the enzyme, even colliding with the substrate.

Answer 95

The enzyme must come in to come physical contact with the substrate and have an active site that fits the substrate. Change rate means one of these factors is affected.

Answer 96

Measuring the time it takes to: - Make a product, e.g. hydrogen peroxide to oxygen and water catalysed by catalase- measure oxygen using gas syringe - Breakdown of product, e.g. the reduction of starch to maltose by amylase- use a spotting tile and iodine and see when turns blue black.

Answer 97

1. At first, lots of substrate and no product (remember to name the substrate and product). 2. It's easy for the substrate to reach an empty active site. **All the active sites are filled/ occupied** and the substrate is rapidly acted upon. Rate of reaction is at its peak. 3. The amount of substrate decreases and the amount of product increases over time, there is less substrate and more product. 4. This makes it more difficult for the substrate to come in contact with the enzyme, as there are fewer and the products get in the way. It takes longer for the substrates to be broken down. The rate slows and the gradient decreases. 5. The rate of reaction becomes so low the changing concentration can't be measured. The graph flats and all the substrate is used up.

Answer 98

The enzyme and substrates names.

Answer 99

Find the gradient of the tangent of the graph. Use the normal to help you find the tangent and draw an equal distance from where it touches the curve.

Answer 100

Only one variable is changed the rest are controlled.

Answer 101

The enzymes are never used up.

Answer 102

1. If the substrate is in excess, an increase in the enzyme is proportional to the rate of reaction increase as there are more substrates than active sites. **Enzymes are more likely to collide with the substrate and form enzyme substrate complexes.** 2. If the substrate is limited, there are more enzymes than substrate. The substrate is the limiting factor, and adding more enzyme has no further effect, as there are empty active sites.

Answer 103

Answer on revision card. Labels should include: * A steady increase as more active sites available. * An increase in enzyme concentration has no further effect.

Answer 104

* At first, a higher substrate concentration increases the rate of reaction proportionally. The more substrate, the **more** likely a collision will occur, and **enzyme substrate complexes** will **form**. * More active sites are occupied until the maximum/ saturation point is reached, at this point, **all active sites are full/ occupied**. * After the saturation point there are more substrates than enzymes. All active sites are full and **enzyme concentration is the limiting factor.** Adding more substrate has no effect on the rate of reactions

Answer 105

**As substrate concentration increases, time to catalyse substrate increases.**

Answer 106

Prevent the enzyme from forming **enzyme substrate complexes.**

Answer 107

* Competitive inhibitors have a molecular **shape similar to the substrate.** * The competitive inhibitors compete with the substrate to **bind with the active site.** * The competitive inhibitor blocks the active site, so no substrate molecules can fit. **Prevents/ reduces enzyme substrate complexes forming.** Reduces amount of substrate catalysed * Not permanent.

Answer 108

* The amount of inhibition is relative to the amount of inhibitor vs. the amount of substrate. * An increased amount of inhibitor means there's less chance of the substrate reaching the active site. * An increased substrate means there's more chance of binding to the active site and the effect of the inhibitor is reduced.

Answer 109

Answer on revision card.

Answer 110

**Bind the enzyme away from the active site** and **alter the shape of** the enzyme and **active site**, **changing the tertiary structure.** The enzyme is **no longer complementary to the substrate** so **substrate can't fit** and the enzyme can't catalyse the reaction. **No enzyme substrate complexes** form.

Answer 111

Answer on revision card.

Answer 112

No effect- the graph shows **increasing the substrate concentration doesn't increase/ change the rate of reaction so it does not overcome inhibition**

Answer 113

* End products of metabolic pathways sometimes act as inhibitors and regulate their own production. * Metabolic pathways are a series of metabolic reactions.

Answer 114

* Co-enzymes improve the rate of reaction * **Bind** to the enzyme away from active site and **alter the tertiary structure** * This **causes the active site to change site** * This makes the active site more complementary to the substrate so **more successful enzyme-substrate complexes form**

Answer 115

-log₁₀H⁺ Approximately equal to standard form.

Answer 116

A measure of hydrogen ion concentration. Inversely proportional to the number of hydrogen ions.

Answer 117

pH where enzymes work fastest. Different for different enzymes.

Answer 118

* At the optimum pH the rate is fastest. * An increase or decrease from the optimum pH e.g. by **adding acid/ alkali** reduces the rate. * A lower or higher pH alters the charges in amino acids in the active site. * The change in pH causes H⁺ and OH^- ions to disrupt the ionic and hydrogen bonds in the enzymes tertiary structure causing them to break and change the active site's tertiary structure. * The enzyme at first finds it difficult to bind to the substrate then cannot bind at all. The enzyme is no longer complementary to the substrate. The substrate can't attatch to the active site and enzyme substrate complexes don't form. The enzyme is **denatured** and the rate decreases to zero.

Answer 119

* Acidic- below pH 7 * Alkaline- above pH 7

Answer 120

Answer on revision card.

Answer 121

* At low temperatures e.g. in fridge/ freezer- **low kinetic energy so fewer enzyme substrate complexes form** * An increased temperature increases the kinetic energy of the enzymes and substrates. Molecules move faster and are more likely to collide and form enzyme substrate complexes. Collisions happen more frequently. The energy of collisions increases- more likely to result in a reaction and enzyme substrate complexes. The graph has a positive gradient. * The enzyme meets the up its optimum temperature where the rate is highest. The graph peaks. * After the optimum, higher temperatures e.g. by **boiling** cause hydrogen and other bonds in the enzyme to break changing the tertiary strucuture and shape of the enzyme and it's active site. The enzyme finds it more difficult to form enzyme substrate complexes, the rate of reaction decreases and the gradient of the graph becomes negative. * Firstly, the substrate doesn't fit as easily and this slows the rate of reaction. After a certain point the enzyme is so changed it stops working and **denatures**. The enzyme is no longer complementary to the substrate and the rate of reaction declines to zero.

Answer 122

Answer on revision card.

Answer 123

- Enzymes optimum temperature is 40°C despite the human body temperature being 35°C. - This difference ensures the safety of the enzyme as any rise in temperature from the optimum could denature the enzyme - This also ensures that less energy and food is needed to gain higher temperature. - It’s also is important to ensure the other proteins do not denature. - This is also the same for bacteria- isolated enzymes can't be tested on their own- **all enzymes/ protiens needed to provide bigger picture**

Answer 124

1. Compare the rate of reaction at the start of the graph and at the end of the graph, and explain the difference. 2. Look at where the maximum is and explain why 3. If given different graphs compare and explain the difference. 4. Identify any effect that denatures the enzyme where reactions stop sooner. 5. Mention why a reaction is stopped Is it because the enzyme has denatured or because the substrate has been used up.

Answer 125

Different for different enzymes

3.1 Chapter 1- Biological Molecules Flashcards

(150 cards)