3.1 Biological Molecules Flashcards

Question

biochemical test using iodine/potassium iodide for starch.

Answer 1

To test for the presence of starch in a sample, add a few drops of orange/brown iodine in potassium iodide solution to the sample The iodine is in potassium iodide solution as iodine is insoluble in water If starch is present, iodide ions in the solution interact with the centre of starch molecules, producing a complex with a distinctive blue-black colour This test is useful in experiments for showing that starch in a sample has been digested by enzymes

Answer 2

Reducing: fructose maltose galactose glucose Non-reducing: sucrose

Answer 3

1) Prepare urine samples Take three test tubes and label each with the name of the patient (urine sample). Add three cm3 urine samples from each patient. To each test tube add 2 cm3 of Benedict’s reagent and mix the contents of the tube. Benedict's reagent is used to detect the presence of glucose. The reagent is blue and turns from blue to red when glucose concentration is greater. The solution also becomes more opaque. 2) Prepare the calibration curve Label six test tubes 0 to 10 mmol dm−3. A glucose solution of 10 mmol dm−3 is provided. Dilute this standard glucose solution using distilled water to produce 5 additional concentrations which will be used to make the calibration curve. The concentrations should be 0 mmol dm−3, 2mmol dm−3, 4mmol dm−3, 6mmol dm−3, 8mmol dm−3, and 10mmol dm−3. 3) Incubate test tubes Add 2 cm3 of Benedict's reagent to each tube and mix the contents. Place all the test tubes into the water bath together (including the tubes with the urine samples) and time for four minutes. Allow to cool before taking readings from the colorimeter. 4) Use the colorimeter Use the contents of the 0.0 mmol dm−3 glucose solution tube, which you have heated with Benedict's, as a blank to calibrate the colorimeter to zero absorbance. Place the remaining samples in cuvettes into the colorimeter and read the absorbance. Record the results

Answer 4

Lipids are: -macromolecules which contain only hydrogen, carbon and oxygen. However, unlike carbohydrates, lipids contain a lower proportion of oxygen -non-polar (electrons have been shared equally so it is not charged) -hydrophobic (insoluble in water) The R-group of a fatty acid may be saturated or unsaturated. In phospholipids, one of the fatty acids of a triglyceride is substituted by a phosphate-containing group.

Answer 5

triglycerides (the main component of fats and oils) and phospholipids

Answer 6

-non-polar, hydrophobic molecules -monomers = glycerol and fatty acids -shorthand chemical formula for a fatty acid is RCOOH Fatty acids can vary in two ways: -Length of the hydrocarbon chain (R group) -The fatty acid chain (R group) may be saturated (mainly in animal fat) or unsaturated (mainly vegetable oils, although there are exceptions e.g. coconut and palm oil) Unsaturated fatty acids can be mono/poly-unsaturated: If H atoms are on the same side of the double bond they are cis-fatty acids and are metabolised by enzymes If H atoms are on opposite sides of the double bond they are trans-fatty acids and cannot form enzyme-substrate complexes, therefore, are not metabolised. They are linked with coronary heart disease MT: -some are liquid at room temperature (oil), whereas others are solid(fat) -glycerol has 3 hydroxyl groups,hence why it needs 3 fatty acids to form a triglyceride -each fatty acid has 1 carboxyl group (-COOH) --the carbon bonds with a hydrocarbon tail to make saturated/unsaturated fatty acids (unsaturated fatty acids can be identified by the presence of a double bond) the hydrocarbon tail is recognised as the R group

Answer 7

Triglycerides are formed by esterification -An ester bond forms when a hydroxyl (-OH) group on glycerol bonds with the carboxyl (-COOH) group of the fatty acid: -An H from glycerol combines with an OH from the fatty acid to make water -The formation of an ester bond is a condensation reaction -For each ester bond formed a water molecule is released -Three fatty acids join to one glycerol molecule to form a triglyceride -Therefore for one triglyceride to form, three water molecules are released

Answer 8

Energy storage -long hydrocarbon chains = little oxygen (triglycerides are highly reduced) -triglycerides are oxidised during cellular respiration = **ester bonds break = energy released = used to produce ATP** -store more energy per gram than carbohydrates and proteins (37kJ compared to 17kJ) -triglycerides are **hydrophobic = don’t cause water uptake in cells by osmosis = more can be stored** -plants store them as oils in seeds and fruits -generally liquid when extracted at room temperature (due to the presence of double bonds which add kinks to the fatty acid chains altering their properties) -mammals **store them as oil droplets in adipose tissue** to help them survive when food is scarce (e.g. hibernating bears) -oxidation of the carbon-hydrogen bonds releases large numbers of water molecules (metabolic water) during cellular respiration -desert animals retain this water if there is no liquid water to drink -bird and reptile embryos in their shells also use this water Insulation -part of the composition of the myelin sheath that surrounds nerve fibres -provides insulation = increased speed of transmission of nerve impulses -also compose part of the adipose tissue layer below the skin -acts as insulation against heat loss (eg. blubber of whales) Buoyancy: -low density of fat tissue = increased ability of animals to float more easily Protection: -adipose tissue in mammals contains stored triglycerides -helps protect organs from the risk of damage

Answer 9

Structure: -two fatty acids bonded to a glycerol molecule AND a phosphate ion (PO43-) -phosphate is polar = soluble in water (hydrophilic) -fatty acid ‘tails’ are non-polar and therefore insoluble in water (hydrophobic) Function: -main component of cell membranes -presence of hydrophobic fatty acid tails creates a hydrophobic core when a phospholipid bilayer forms -acts as a barrier to water-soluble molecules -hydrophilic phosphate heads form H-bonds with water -allows cell membrane to be used to compartmentalise -enables cells to organise specific roles into organelles helping with efficiency -composition of phospholipids contributes to fluidity of the cell membrane -mainly saturated fatty acid tails = membrane will be less fluid -mainly unsaturated fatty acid tails = membrane will be more fluid Control membrane protein orientation: -weak hydrophobic interactions between the phospholipids and membrane proteins hold the proteins within the membrane but still allow movement within the layer -when water is present, phospholipids will form monolayers or bilayers, due to the hydrophobic and hydrophilic parts,

Answer 10

Amino acids

Answer 11

Carbon in the centre A hydrogen atom NH2 to represent an amine group COOH t represent a carboxyl group R group represents a side chain

Answer 12

-condensation reaction occurs between amino acids to form a peptide bond -1 water molecule released in this process -reverse reaction occurs during digestion -dipeptides are formed by the condensation of two amino acids -polypeptides are formed by the condensation of many amino acids -sequence of amino acids in a polypeptide is determined by the sequence of the genetic code on mRNA being translated in the ribosomes

Answer 13

Primary structure: -unique sequence of amino acids in polypeptide chain Primary structure proteins vary depending on 3 factors: -number of amino acids -sequence of amino acids -different R groups Secondary structure: -primary structure folds -held together by hydrogen bonds between the carboxyl groups and amino groups in the polypeptide backbone -can folded in two ways: -alpha-helix, a tight, twisted strand -beta pleated sheet, a zig-zag line of amino acids bonds with the next to form a ribbon shape Tertiary structure: -further folding of the secondary structure -held together by bonds between the R groups of the amino acids in the protein, so depends on the sequence of amino acids -3 types of bonds involved: -weak hydrogen bonds -ionic bonds between R-groups with positive/negative charges -disulfide bridges - covalent S-S bonds between two cysteine amino acids which are the strongest type of bond, allowing it to withstand higher temperatures and extreme pH (Note: disulfide bonds are not present in every tertiary structure: only when R groups contain sulfur) Quaternary structure: -formed from several polypeptides aka subunits -common examples include: -collagen - fibrous protein of three polypeptide chains (trimetric) -haemoglobin - globular protein with four polypeptide chains (tetrametric) -insulin - globular: two polypeptide chains (dimetric) held together by disulfide bond Tertiary and quaternary structure proteins can be globular or fibrous: Globular proteins: -spherical -hydrophilic amino acids on the outside, allowing them to interact with aqueous environment -hydrophobic amino acids shielded on the inside of the protein, as they are non-polar -hence globular proteins are soluble in water Examples of functional proteins: -plasma proteins in bloodstream -enzymes are able to catalyse reactions due to their solubility -haemoglobin in RBCs transport oxygen and contains iron as a prosthetic group (inorganic compound) Fibrous proteins: -strands -both hydrophobic and hydrophilic are exposed to the outside, hence making it insoluble and ideal for structural purposes Examples of structural proteins: -collagen (connective tissue) -elastin (found in alveoli and blood vessels)

Answer 14

Method: 1. Take 2 ml of the solution to be tested in a test tube 2. Add 2 ml of 5% sodium hydroxide solution to make the sample alkaline 3. Mix the solutions 4. Add two drops of 1% copper sulfate solution 5. A colour change from blue to violet indicates the presence of proteins How it works: The Cu2+ ions react with the nitrogen of the peptide bond to form a purple colour

Answer 15

-biological catalyst (speeds up reaction without being used up) -globular protein -soluble in water, so can move freely in organelles and cytoplasm

Answer 16

-due to complementary nature between the shape of the active site on the enzyme and its substrate(s) -shape of active site (and hence the specificity of the enzyme) is determined by the complex tertiary structure of the protein that makes up the enzyme: -proteins are formed from chains of amino acids held together by peptide bonds -order of amino acids determines the shape of an enzyme -if the order is altered, the resulting three-dimensional shape changes -enzyme-substrate complex forms when an enzyme and its substrate join together, but only temporarily, before the enzyme catalyses the reaction and the product(s) are released

Answer 17

-activation energy = amount of energy needed by substrate to become just unstable enough for a reaction to occur and for products to form -enzymes speed up chemical reactions because they influence the stability of bonds in the reactants -the destabilisation of bonds in the substrate makes it more reactive -enzymes work by lowering the activation energy of a reaction and in doing so they provide an alternative energy pathway

Answer 18

Catabolic reactions -involve the breakdown of complex molecules into simpler products, which happens when a single substrate is drawn into the active site and broken apart into two or more distinct molecules e.g. cellular respiration and hydrolysis reactions Anabolic reactions -involve the building of more complex molecules from simpler ones by drawing two or more substrates into the active site, forming bonds between them and releasing a single product e.g. protein synthesis and photosynthesis

Answer 19

-enzymes are globular proteins, so their shape (as well as the shape of the active site) is determined by the complex tertiary structure of the protein that makes up the enzyme -therefore highly specific -first model of enzyme activity -suggested that both enzymes and substrates were rigid structures that locked into each other very precisely -later modified and adapted to our current understanding of enzyme activity, permitted by advances in techniques in the molecular sciences

Answer 20

-scientists discovered that proteins are not rigid structures -found that multiple regions of an enzyme molecule moved in response to the environment -larger movements occurred when the substrate bound to the enzyme -prior to binding, the substrate and active site and not completely complementary in shape -when the substrate binds the active site alters shape and moulds around the substrate (known as conformational changes) -maximises the ability of the enzyme to catalyse the reaction

Answer 21

-the higher the enzyme concentration in a reaction mixture, the greater the number of active sites available and the greater the likelihood of enzyme-substrate complex formation -as long as there is sufficient substrate available, the initial rate of reaction increases linearly with enzyme concentration -if the amount of substrate is limited, at a certain point any further increase in enzyme concentration will not increase the reaction rate as the amount of substrate becomes a limiting factor

Answer 22

-the greater the substrate concentration, the higher the rate of reaction -as the number of substrate molecules increases, the likelihood of enzyme-substrate complex formation increases -if the enzyme concentration remains fixed but the amount of substrate is increased past a certain point, however, all available active sites eventually become saturated and any further increase in substrate concentration will not increase the reaction rate -when the active sites are all full, any substrate molecules that are added have nowhere to bind to form an enzyme-substrate complex

Answer 23

two types of reversible inhibitors: -competitive inhibitors have a similar shape to that of the substrate molecules and therefore compete with the substrate for the active site -non-competitive inhibitors bind to the enzyme at an alternative site, which alters the shape of the active site and therefore prevents the substrate from binding to it -slow down reactions (opposite of catalyst) Competitive inhibitors: -active site, substrate and inhibitors all have similar shape -means that inhibitors bind to active site, which blocks the substrate from binding to active site -hence less E-S complexes are made, so less products are formed per unit time, so rate of reaction decreases -tend to be reversible, so bonds are temporary and active site is not changed, meaning the enzyme can still function as a catalyst, just not as effectively Non-competitive inhibitors: -do not have the same shape as active site -therefore, inhibitor binds to any part of the enzyme, away from active site (known as the allosteric site) -this alters shape of protein, due to enzyme flexibility (proven by induced fit model, but not lock and key) -therefore, change in active site, so it is no longer complementary to substrate -hence less E-S complexes are made, so less products are formed per unit time, so rate of reaction decreases -tend to be irreversible, due to permanent bond between allosteric site and inhibitor

Answer 24

-enzymes have a specific optimum temperature -lower temperatures either prevent reactions from proceeding or slow them down: -molecules move relatively slow -lower frequency of successful collisions between substrate molecules and active site of enzyme -frequent enzyme-substrate complex formation -substrate and enzyme collide with less energy, making it less likely for bonds to be formed or broken (stopping the reaction from occurring) -higher temperatures speed up reactions: -molecules move more quickly -higher frequency successful collisions between substrate molecules and active site of enzyme -more frequent enzyme-substrate complex formation -substrate and enzyme collide with more energy, making it more likely for bonds to be formed or broken (allowing the reaction to occur) However, as temperatures continue to increase, the rate at which an enzyme catalyses a reaction drops sharply, as the enzyme begins to denature: -bonds (eg. hydrogen bonds) holding the enzyme molecule in its precise shape start to break -causes the tertiary structure of the protein (ie. the enzyme) to change -permanently damages the active site, preventing the substrate from binding -denaturation has occurred if the substrate can no longer bind -few human enzymes can function at temperatures above 50°C -due to humans maintaining a body temperature of about 37°C, therefore even temperatures exceeding 40°C will cause the denaturation of enzymes -high temperatures causes hydrogen bonds between amino acids to break, changing the conformation of the enzyme

Answer 25

-enzymes are denatured at extremes of pH -hydrogen and ionic bonds hold the tertiary structure of the protein (ie. the enzyme) together -below and above the optimum pH of an enzyme, solutions with an excess of H+ ions (acidic solutions) and OH- ions (alkaline solutions) can cause these bonds to break -alters the shape of the active site, which means enzyme-substrate complexes form less easily -eventually, enzyme-substrate complexes can no longer form at all, so complete denaturation of the enzyme has occurred Where an enzyme functions can be an indicator of its optimal environment: -Eg. pepsin is found in the stomach, an acidic environment at pH 2 (due to the presence of hydrochloric acid in the stomach’s gastric juice) -Pepsin’s optimum pH, not surprisingly, is pH 2

Answer 26

Generally: A pentose sugar (a sugar with 5 carbon atoms) A nitrogen-containing organic base A phosphate group DNA -1 deoxyribose sugar with hydrogen at the 2' position -a phosphate group -a nitrogenous base either adenine (A), cytosine(C), guanine(G) or thymine(T) -by complementary base pairing, hydrogen bonds form between bases -A and T have 2 hydrogen bonds -C and G have 3 hydrogen bonds RNA -a ribose sugar with a hydroxyl (OH) group at the 2' position -a phosphate group -a nitrogenous base either adenine (A), cytosine(C), guanine(G) or uracil(U)

Answer 27

-DNA holds genetic information -contains the instructions for the growth and development of all organisms -RNA transfers genetic information from DNA to the ribosomes, which read RNA to synthesise proteins in translation -both needed to build proteins, which are essential for the proper functioning of cells

Answer 28

-nitrogenous base molecules that are found in DNA (A, T, C, G) and RNA (A, U, C, G) occur in two structural forms: purines and pyramidines -adenine and guanine are purines – they have a double ring structure -cytosine, thymine and uracil are pyrimidines – they have a single ring structure

Answer 29

DNA -double helix with two polynucleotide chains held together by hydrogen bonds between specific complementary base pairs -antiparallel -each DNA polynucleotide strand has a 3’ end and a 5’ end (these numbers relate to which carbon on the pentose sugar could be bonded with another nucleotide) -as the strands run in opposite directions, one is known as the 5’ to 3’ strand and the other is known as the 3’ to 5’ strand RNA -relatively short polynucleotide chain -only made of one strand -consists of alternating ribose sugars and phosphate groups linked together The sugar-phosphate bonds (between different nucleotides in the same strand) are covalent / phosphodiester bonds -form sugar-phosphate backbone of the RNA polynucleotide strand -phosphodiester bonds link the 5-carbon of one ribose sugar molecule to phosphate group from same nucleotide, which is linked by another phosphodiester bond to the 3-carbon of the ribose sugar molecule of the next nucleotide in the strand -both are polynucleotides -nucleotides are joined via condensation reactions between phosphate group of one nucleotide and pentose sugar of the next nucleotide -forms phosphodiester bond as it consists of a phosphate group and two ester bonds -sugar phosphate backbone = chain of alternating phosphate groups and pentose sugars produced as a result of many phosphodiester bonds

Answer 30

ensures genetic continuity between generations of cells

Answer 31

-in each new DNA molecule produced, one of the polynucleotide DNA strands is from the original DNA molecule being copied -the other polynucleotide DNA strand has to be newly created by the cell -DNA Helicase breaks hydrogen bonds between the complementary bases -double strand separates, leaves 2 template stands -free complementary nucleotides are attracted to and bind to exposed bases on template strands (A to T, C to G) -DNA Polymerase joins the sugar-phosphate backbone of the new strand

Answer 32

-Replicating Bacterial DNA in 2 types of Nitrogen Isotopes, 15N and 14N -15N = heavy isotope -14N = light isotope -Nitrogen found in nitrogenous bases of DNA -Bacterial DNA made from 15N will have a Heavy Density -Bacterial DNA made from 14N will have a Light Density -Experiment = Bacterial DNA made of 15N is replicated in an environment of 14N – produces DNA molecules with half 15/half 14 (semi-conservative replication, original strand = 15N & new strand = 14N), therefore, DNA molecule has medium density

Answer 33

-metabolite in many metabolic reactions, including condensation/hydrolysis reactions -solvent in which metabolic reactions occur -has relatively high heat capacity, buffers changes in temperature -has relatively large latent heat of vaporisation, providing a cooling effect with little loss of water through evaporation -has strong cohesion between water molecules, which supports columns of water in the xylem and produces surface tension where water meets air

Answer 34

a molecule of ribose, a molecule of adenine and three phosphate groups Hydrolysis: ATP —> ADP + Pi (catalysed by ATP hydrolase) -inorganic phosphate released during the hydrolysis of ATP can be used to phosphorylate other compounds -often makes them more reactive Resynthesis ADP + Pi —> ATP (catalysed by ATP synthase)

3.1 Biological Molecules Flashcards

(59 cards)