Gatika's A2 Biology Revision Guide PDF

INTERNATONAL A-LEVEL FOR CIE A2 BIOLOGY REVISION GUIDE  GATIKA’S VERSION OF A2 BIOLOGY Page number Chapter 1: Energy & Respiration 3-9 Chapter 2: Photosynthesis...

INTERNATONAL A-LEVEL FOR CIE A2 BIOLOGY REVISION GUIDE  GATIKA’S VERSION OF A2 BIOLOGY Page number Chapter 1: Energy & Respiration 3-9 Chapter 2: Photosynthesis 10-16 Chapter 3: Homeostasis 17-23 Chapter 4: Control & Coordination 24-32 Chapter 5: Inheritance 33-39 Chapter 6: Selection & Evolution 40-50 Chapter 7: Classification, Biodiversity & Conservation 51-63 Chapter 8: Genetic Technology 64-78 CHAPTER 1 : ENERGY AND RESPIRATION Q1. Examples of energy requiring processes in the body. Transporting substances across membrane including active transport using sodium-potassium pumps in cell membrane and exocytosis of digested bacteria from WBCs. Anabolic reactions including DNA synthesis from nucleotides and protein synthesis from amino acids. Movement including cellular movement of chromosomes via spindle and mechanical contraction of muscles. Maintaining body temperature. Q2. Respiration equation in animals. C6H12O6 + 6O2 -> 6CO2 + 6H2O Q3. Structure of ATP. ATP is a phosphorylated nucleotide. It is made of: ribose sugar, adenine base, phosphate groups Adenosine monophosphate: adenosine + 1 phosphate Adenosine diphosphate: adenosine + 2 phosphate Adenosine triphosphate: adenosine + 3 phosphate (adenine + ribose = adenosine) Q4. Features of ATP which make it suitable as universal energy currency.  Hydrolysis of ATP can be carried out quickly and easily wherever energy is required within cell by action of just one enzyme ATPase  A useful quantity of energy is released from hydrolysis of one ATP molecule  Relatively stable at cellular pH values  Can be recycled, breakdown of ATP is reversible reaction; ATP can be formed from ADP and Pi  Soluble and moves easily within cell, so it can transport energy to different areas of cell  Forms phosphorylated intermediates, this can make metabolites more reaction & lower activation energy required for reaction Q5. Two types of ATP synthesis. 1. Substrate linked phosphorylation  ATP is formed by transferring a phosphate directly from substrate molecule to ADP ADP + Pi -> ATP  Energy required for the reaction is provided directly by another chemical reaction  Occurs in cell cytoplasm and matrix of mitochondria  Only accounts for small amount of ATP synthesized during aerobic respiration  Takes place in glycolysis 2. Chemiosmosis  Involves a proton (hydrogen ion) gradient across membrane  Takes place across inner membrane of mitochondria & thylakoid membrane of chloroplast  An electron transport chain helps to establish proton concentration gradient   high energy electrons move from carrier to carrier releasing energy that is used to pump protons (up a concentration gradient) across inner membrane into intermembrane space  protons are pumped from a low concentration in mitochondrial matrix to high concentration in intermembrane space  Protons then move down concentration gradient into matrix which releases energy  Protons move through ATP synthase complex which uses released energy to drive phosphorylation of ATP  Oxygen acts as final electron & proton acceptor to form water  Most of ATP made during respiration is synthesized via chemiosmosis Q6. Energy values of different respiratory substrate. Respiratory substrate Energy value kJ/g Carbohydrate 15.8 Lipid 39.4 Protein 17.0 Q7. Differences in energy values of different substrates. Differences in energy values of substrates can be explained by their molecular composition, specifically how many hydrogen atoms become available when substrate molecule is broken down.  Substrate molecules are broken down & hydrogen atoms become available  Hydrogen carrier molecules (NAD & FAD) pick them up (become reduced) & transfer them to inner mitochondrial membrane  Reduced NAD and FAD release hydrogen atom which split into protons & electrons  Protons are pumped across inner mitochondrial membrane into the intermembrane space, forming a proton/chemiosmotic gradient  Proton gradient is used in chemiosmosis to produce ATP  After proton have flowed back into matrix of mitochondria via ATP synthase, they are oxidised to form water This means that a molecule with higher hydrogen content will result in greater proton gradient across the mitochondrial membrane which allows for formation of more ATP via chemiosmosis. Fatty acids in lipids are made up of long hydrocarbon chains with lots of hydrogen atoms. These hydrogen atoms are released when lipid is broken down. Q8. What is RQ and RQ values of different respiratory substrates. Respiratory quotient (RQ) is the ratio of CO2 molecules produced to oxygen taken in during respiration. RQ = moles or molecules of CO2 ÷ moles or molecules of O2 Respiratory Substrate RQ Carbohydrate 1.0 Lipid 0.7 Protein 0.9 Q9. Structure and function of mitochondria.  Mitochondria are rod-shaped organelle; 0.5-1 μm in diameter   They are site of aerobic respiration and their function is to synthesis ATP  Mitochondria have two phospholipid membrane  The outer membrane is smooth and permeable to several small molecules.  Inner membrane is folded to form cristae, less permeable, site of electron transport chain and location of ATP synthase  Intermembrane space has a low pH due to high concentration of protons; Concentration gradient across membrane is formed during oxidative phosphorylation and it is essential for ATP synthase  Matrix is an aqueous solution within inner membrane of mitochondrion; It contains ribosomes, enzymes & circular mitochondrial DNA Q10. Adaptations of inner mitochondrial membrane. The inner mitochondrial membrane has a large surface, due to the presence of inner folds called cristae. This enables the membrane to hold many electron transport chain proteins & ATP synthase enzymes. The electron transport chain is used to generate an electrochemical gradient across the membrane, it drives ATP synthesis. Protons diffuse back via ATP synthase releasing energy, which is used to synthesise ATP. Q11. Stages involved in aerobic respiration and their location in the cell. STAGE LOCATION Glycolysis Cell cytoplasm Link Reaction Matrix of mitochondria Kreb Cycle Matrix of mitochondria Oxidative Phosphorylation Inner membrane of mitochondria Q12. Steps involved in glycolysis. Phosphorylation: glucose (6C) is phosphorylated by 2 ATP to form fructose biphosphate (6C) Glucose + 2ATP -> Fructose biphospate Lysis: fructose biphosphate (6C) splits into two molecules of triose phosphate (3C) Fructose biphosphate -> 2 Triose phosphate Oxidation: hydrogen is removed from each molecule of triose phosphate and transferred to coenzyme NAD to form 2 reduced NAD 4H + 2NAD -> 2NADH + 2H+ Dephosphorylation: phosphates are transferred from the intermediate substrate molecules to form 4 ATP through substrate-linked phosphorylation 4Pi + 4ADP -> 4ATP Pyruvate produced: end product of glycolysis which can be used in the next stage of respiration 2 Triose phosphate -> 2 Pyruvate Q13. Link reaction and the role of coenzyme A. Link reaction links glycolysis to the Krebs cycle. It involves two steps:  1. Decarboxylation and dehydrogenation of pyruvate by enzymes to produce an acetyl group, CH3C(O)- 2. Combination with coenzyme A to form acetyl CoA Pyruvate + NAD + CoA -> acetyl CoA + carbon dioxide + reduced NAD End products of glycolysis are acetyl CoA, carbon dioxide (CO2) and reduced NAD (NADH). Coenzyme A consists of a nucleoside (ribose and adenine) and a vitamin. In the link reaction, CoA binds to the remainder of the pyruvate molecule (acetyl group 2C) to form acetyl CoA. It then supplies the acetyl group to the Krebs cycle where it is used to continue aerobic respiration. Q14. Steps involved in Kreb Cycle.  Acetyl CoA (2C) enters the circular pathway via the link reaction  4 carbon (4C) oxaloacetate accepts the 2C acetyl fragment from acetyl CoA to form citrate (6C)  Oxaloacetate is regenerated via series of reaction: o Decarboxylation of citrate - releasing 2 CO2 as waste gas o Dehydrogenation of citrate - releasing H atoms that reduce coenzymes NAD and FAD 8H + 3NAD + FAD -> 3NADH + 3H+ + FADH2 o Substrate-level phosphorylation - a phosphate is transferred from one of the intermediates to ADP, forming 1 ATP Q15. Role of NAD and FAD in respiration. When hydrogen atoms become available at different points during respiration NAD and FAD accept these hydrogen atoms. A hydrogen atom consists of a hydrogen ion and an electron. When the coenzymes gain a hydrogen they are reduced. They transfer the hydrogen atoms (hydrogen ions and electrons) from the different stages of respiration to the electron transport chain on the inner mitochondrial membrane, the site where hydrogens are removed from the coenzymes. When the hydrogen atoms are removed the coenzymes are oxidised. Hydrogen ions and electrons are important in the electron transport chain at the end of respiration as they play a role in the synthesis of ATP  Electrons from reduced NAD (NADH) and reduced FAD (FADH2) are given to the electron transport chain  Hydrogen ions from reduced NAD (NADH) and reduced FAD (FADH2) are released when the electrons are lost  The electron transport chain drives the movement of these hydrogen ions (protons) across the inner mitochondrial membrane into the intermembrane space, creating a proton gradient (more hydrogen ions in the intermembrane space)  Movement of hydrogen ions down the proton gradient, back into the mitochondrial matrix, gives the energy required for ATP synthesis Q16. What is oxidative phosphorylation. Oxidative phosphorylation is the last stage of aerobic respiration. It takes place at the inner membrane of the mitochondria.  Hydrogen atoms are donated by reduced NAD and FAD  Hydrogen atoms split into protons and electrons  The high energy electrons release energy as they move through the electron transport chain  The released energy is used to transport protons across the inner mitochondrial membrane from the matrix into the intermembrane space  A concentration gradient of protons is established between the intermembrane space and the matrix  The protons return to the matrix via facilitated diffusion through the channel protein ATP synthase  The movement of protons down their concentration gradient provides energy for ATP synthesis  Oxygen combines with protons and electrons at the end of the electron transport chain to form water Q17. Ethanol fermentation.  Reduced NAD transfers its hydrogens to ethanal to form ethanol  In the first step of the pathway pyruvate is decarboxylated to ethanal, producing CO2  Then ethanal is reduced to ethanol by the enzyme alcohol dehydrogenase  Ethanal is the hydrogen acceptor  Ethanol cannot be further metabolised; it is a waste product Q18. Lactate fermentation.  Reduced NAD transfers its hydrogens to pyruvate to form lactate  Pyruvate is reduced to lactate by enzyme lactate dehydrogenase  Pyruvate is the hydrogen acceptor  The final product lactate can be further metabolised Q19. Further metabolization of lactate. After lactate is produced two things can happen: 1. It can be oxidised back to pyruvate which is then channelled into the Krebs cycle for ATP production 2. It can be converted into glycogen for storage in the liver The oxidation of lactate back to pyruvate needs extra oxygen. This extra oxygen is referred to as an oxygen debt. Q20. Energy yield: aerobic vs. anaerobic In cells there is a much greater energy yield from respiration in aerobic conditions than in anaerobic conditions. In anaerobic respiration glucose is only partially oxidised meaning only some of its chemical potential energy is released and transferred to ATP. The only ATP producing reaction that continues is glycolysis (~2 ATP). As there is no oxygen to act as the final electron acceptor none of the reactions within the mitochondria can take place. The stages that take place inside the mitochondria produce much more ATP than glycolysis alone (~36 ATP). Q21. Adaptations of rice plants for aerobic and anaerobic respiration. Aerobic respiration: Some types of rice show an increased rate of upward growth away from the waterline. The leaves always remain above water so there is access to oxygen and carbon dioxide through the stomata. Rice plants possess aerenchyma tissue in the stems and roots. This specialised plant tissue contains useful air spaces that allow gases that enter the stomata to diffuse to other parts of the plant that are above and under the water. Oxygen and carbon dioxide can therefore be held in this tissue even when underwater and can be transferred from parts of the plant that has access to air. Anaerobic respiration: Plants use ethanol fermentation during anaerobic respiration. Toxic ethanol is produced which can build up in the plant tissue causing damage. Rice plants can tolerate higher levels of toxic ethanol compared to other plants. They also produce more ethanol dehydrogenase. This is the enzyme that breaks down ethanol. The resilience that rice plants have towards ethanol allows them to carry out anaerobic respiration for longer so enough ATP is produced for the plant to survive and actively grow. Q22. Investigating effect of temperature and substrate concentration on aerobic respiration. Dehydrogenation happens regularly throughout the different stages of aerobic respiration. The hydrogens that are removed from substrate molecules are transferred to the final stage of aerobic respiration, oxidative phosphorylation, via the hydrogen carriers NAD and FAD. When DCPIP and methylene blue are present they can also take up hydrogens and get reduced. Both redox indicators undergo the same colour change when they are reduced [Blue → colourless]. The faster the rate of respiration, the faster the rate of hydrogen release and the faster the dyes get reduced and change colour. This means that the rate of colour change can correspond to the rate of respiration in yeast. The rate of respiration is inversely proportional to the time taken.  The effect of temperature can be investigated by adding the test tubes containing the yeast suspension to a temperature-controlled water bath and recording the time taken for a colour change to occur once the dye is added. Repeat across a range of temperatures. For example, 30oC, 35oC, 40oC, 45oC.  The effect of substrate concentration can be investigated by adding different concentrations of a substrate to the suspension of yeast cells and recording the time taken for a colour change to occur once the dye is added. For example, 0.1% glucose, 0.5% glucose, 1.0% glucose. Q23. What are respirometers and why are they used? Respirometers are used to measure and investigate the rate of oxygen consumption during aerobic respiration in organisms. By adding the apparatus to a thermostatically controlled water bath the effect of temperature on the rate of respiration can be investigated. The experiments usually involve organisms such as seeds or invertebrates Method:  Measure oxygen consumption: set up the respirometer and run the experiment with both tubes in a controlled temperature water bath. Use the manometer reading to calculate the change in gas volume within a given time, x cm3 min-1  Reset the apparatus: Allow air to reenter the tubes via the screw cap and reset the manometer fluid using the syringe. Change the temperature of the water bath and allow the tubes to acclimate, then close the screw clip to begin the experiment  Run the experiment again: use the manometer reading to calculate the change in gas volume in a given time, y cm3 min-1  Repeat experiment several times at different temperatures Calculations: The volume of oxygen consumed (cm3 min-1) can be worked out using the diameter of the capillary tube r (cm) and the distance moved by the manometer fluid h (cm) in a minute using the formula: πr2h Analysis:  The rate of oxygen consumption (cm3 min-1) is often taken as the rate of respiration for organisms.  The different volumes of oxygen consumed obtained for the different temperatures can be presented in table or graph form to show the effects of temperature. CHAPTER 2 : PHOTOSYNTHESIS Q1. Describe the chloroplast structures and their functions.  Chloroplast are organelles in plant cell where photosynthesis occurs.  Each chloroplast is surrounded by a double-membrane envelope. Each of them membrane envelope is phospholipid bilayer.  Chloroplasts are filled with fluid known as stroma. Stroma is the site of light-independent stage of photosynthesis. Stroma contains 70S ribosomes, loop of DNA & starch grains. Loop of DNA codes for some of chloroplast proteins and those which are coded by this loop are produced at the 70S ribosomes. Sugars formed during photosynthesis are stored as starch inside starch grains.  A separate membrane system is found in stroma:  This is the site of light-dependent stage of photosynthesis.  Membrane contains pigments, enzymes & electron carriers for light-dependent reactions reactions.  Membrane system consists of series of flattened fluid-filled sacs known as thylakoids.  Thylakoids stack up forming structures known as grana.  Grana are connected by membranous channel – stroma lamellae.  Membrane of grana create a large surface area to increase number of light-dependent reactions that can occur.  Membrane system provides large number of pigment molecules in an arrangement that ensures as much light as necessary is absorbed. Q2. Adaptations for photosynthesis.  Plant cells contain chloroplasts which is the site of photosynthesis  Chloroplasts are filled with a fluid known as the stroma  The system of membranes found in the stroma of the chloroplast consists of a series of flattened fluid-filled sacs known as thylakoids  In places, these thylakoids stack up to form structures known as grana (singular – granum)  The light-dependent stage of photosynthesis occurs in the thylakoid membranes and the thylakoid spaces (the spaces inside the thylakoids)  The thylakoid membranes contain the pigments, enzymes and electron carriers required for the light-dependent reactions  The membranes of the grana create a large surface area to increase the number of light dependent reactions that can occur  This membrane system provides a large number of pigment molecules in an arrangement that ensures as much light as necessary is absorbed  The pigment molecules are arranged in light-harvesting clusters known as photosystems  In a photosystem, the different pigment molecules are arranged in funnel-like structures the thylakoid membrane (each pigment molecule passes energy down to the next pigment molecule in the cluster until it reaches the primary pigment reaction centre) Q3. What are the two stages involved in photosynthesis. Light-dependent stage and light-independent stage (Calvin Cycle) Light dependent stage:  Reduced NADP is produced when hydrogen ions combine with the carrier molecule  NADP using electrons from the photolysis of water  ATP is produced (from ADP and Pi by ATP synthase in a process called photophosphorylation (ADP + Pi → ATP)  Photophosphorylation uses the proton (H+) gradient generated by the photolysis of water  Energy from ATP and hydrogen from reduced NADP are passed from the light dependent stage to the light-independent stage of photosynthesis Q4. Describe the role of photosynthetic pigments. Role of photosynthetic pigments is to absorb light energy, to excite electrons for cyclic and non- cyclic photophosphorylation. In a photosystem, the different pigment molecules are arranged in funnel-like structures in the thylakoid membrane (each pigment molecule passes energy down to the next pigment molecule in the cluster until it reaches the primary pigment reaction centre). There are two groups of pigments: Chlorophylls: chlorophyll a and b (chlorophylls absorb wavelength in blue-violet and red regions of light spectra Carotenoids: carotene and xanthophyll (carotenoids absorb mainly in blue-violet region of spectra) Chlorophyll a and b are primary pigments. Carotenoids and xanthophylls are accessory pigments. They pass the energy to the primary pigment reaction centre. Q5. What is an absorption spectra and what does it show.  An absorption spectrum is a graph that shows the absorbance of different wavelengths of light by a particular pigment  Chlorophylls absorb wavelengths in the blue-violet and red regions of the light spectrum  Carotenoids absorb wavelengths of light mainly in the blue-violet region of the spectrum Q6. What is an action spectra and what does it show.  An action spectrum is a graph that shows the rate of photosynthesis at different wavelengths of light  The rate of photosynthesis is highest at the blue- violet and red regions of the light spectrum, as these are the wavelengths of light that plants can absorb (ie. the wavelengths of light that chlorophylls and carotenoids can absorb) Q7. Commonly used technique to separate the chloroplast pigments.  Chromatography is an experimental technique that is used to separate mixtures: o The mixture is dissolved in a fluid/solvent (called the mobile phase) and the dissolved mixture then passes through a static material (called the stationary phase) o Different components within the mixture travel through the material at different speeds o This causes the different components to separate  A retardation factor (Rf) can be calculated for each component of the mixture Rf value = distance travelled by component ÷ distance travelled by solvent  Two of the most common techniques for separating these photosynthetic pigments are: o Paper chromatography – the mixture of pigments is passed through paper (cellulose) o Thin-layer chromatography – the mixture of pigments is passed through a thin layer of adsorbent (eg. silica gel), through which the mixture travels faster and separates more distinctly  Chromatography can be used to separate and identify chloroplast pigments that have been extracted from a leaf as each pigment will have a unique Rf value  The Rf value demonstrates how far a dissolved pigment travels through the stationary phase o A smaller Rf value indicates the pigment is less soluble and larger in size Although specific Rf values depend on the solvent that is being used, in general:  Carotenoids have the highest Rf values (usually close to 1)  Chlorophyll B has a much lower Rf value  Chlorophyll A has an Rf value somewhere between those of carotenoids and chlorophyll B  Small Rf values indicate the pigment is less soluble and larger in size Q8. Describe cyclic photophosphorylation. Cyclic photophosphorylation involves photosystem I (PSI) only  Light is absorbed by photosystem I (located in the thylakoid membrane) and passed to the photosystem I primary pigment (P700)  An electron in the primary pigment molecule (ie. the chlorophyll molecule) is excited to a higher energy level and is emitted from the chlorophyll molecule in a process known as photoactivation  This excited electron is captured by an electron acceptor, transported via a chain of electron carriers known as an electron transport chain before being passed back to the chlorophyll molecule in photosystem I (hence: cyclic)  As electrons pass through the electron transport chain they provide energy to transport protons (H+) from the stroma to the thylakoid lumen via a proton pump  A build-up of protons in the thylakoid lumen can then be used to drive the synthesis of  ATP from ADP and an inorganic phosphate group (Pi) by the process of chemiosmosis Chemiosmosis is the movement of chemicals (protons) down their concentration gradient, the energy released from this can be used by ATP synthase to synthesise ATP  The ATP then passes to the light-independent reactions Q9. Describe non-cyclic photophosphorylation. Non-cyclic photophosphorylation involves both photosystem I (PSI) and photosystem II (PSII) Photosystem II  Light is absorbed by photosystem II (located in the thylakoid membrane) and passed to the photosystem II primary pigment (P680)  An electron in the primary pigment molecule (ie. the chlorophyll molecule) is excited to a higher energy level and is emitted from the chlorophyll molecule in a process known as photoactivation  This excited electron is passed down a chain of electron carriers known as an electron transport chain, before being passed on to photosystem I  During this process to ATP is synthesised from ADP and an inorganic phosphate group (Pi) by the process of chemiosmosis  The ATP then passes to the light-independent reactions  Photosystem II contains a water-splitting enzyme called the oxygen-evolving complex which catalyses the breakdown (photolysis) of water by light:  H2O → 2H+ + 2e- + ½O2  As the excited electrons leave the primary pigment of photosystem II and are passed on to photosystem I, they are replaced by electrons from the photolysis of water Photosystem I  At the same time as photoactivation of electrons in photosystem II, electrons in photosystem I also undergo photoactivation  The excited electrons from photosystem I also pass along an electron transport chain  These electrons combine with hydrogen ions (produced by the photolysis of water) and the carrier molecule NADP to give reduced NADP: 2H+ + 2e- + NADP → reduced NADP  The reduced NADP (NADPH) then passes to the light-independent reactions to be used in the synthesis of carbohydrate Q10. Distinguish between photosystem I (PSI) and photosystem II (PSII). PHOTOSYSTEM I PHOTOSYSTEM II Located on outer surface of thylakoid membrane Located on inner surface of thylakoid membrane Photocenter is P700 Photocenter is P680 Pigments absorb longer wavelengths of light Pigments absorb shorter wavelengths of light (>680 nm) ( receptor -> sensory neurone -> relay neurone -> motor neurone -> effector -> response Q6. Sequence of events resulting in an action potential (tasting salt)  Sodium ions diffuse through highly selective channel proteins in cell surface membrane of microvilli of chemoreceptor cells  This leads to depolarisation of chemoreceptor cell membrane  Increase in positive charge inside cell is known as receptor potential  If there is sufficient stimulation, and depolarisation of membrane, receptor potential becomes large enough to stimulate voltage-gated calcium ion channel proteins to open  Calcium ions enter the cytoplasm of chemoreceptor cell & stimulate exocytosis of vesicles containing neurotransmitter from basal membrane of chemoreceptor  Neurotransmitter stimulates an action potential in sensory neurone  Sensory neurone then transmits an impulse to brain Q7. Describe the relationship between the strength of the stimulus and the resulting action potential. When receptors are stimulated, they are depolarized. If stimulus is very weak or below a certain threshold, receptor cells won’t be sufficiently depolarised, and sensory neurone will not send an impulse. If stimulus is strong enough to increase receptor potential above the threshold potential then receptor will stimulated sensory neurone to send impulse. So, an impulse is only transmitted if initial stimulus is sufficient to increase membrane potential above a threshold potential. The speed at which action potential travels does not vary according to the size of stimulus. Speed of impulse transmission is always the same. Difference is the frequency of the impulse. Strong stimulus produces a rapid succession of action potential, one after the other. Whereas, a weak stimulus results in fewer action potentials per second. This is because, strong stimulus is likely to stimulate more neurons than a weak stimulus. Q8. Describe how a resting potential is set up and maintained in a myelinated neurone. In a resting axon, inside of axon always has a slightly negative electric potential compared to the outside. The potential difference is usually about -70mV (inside of axon has an electric potential 70mV lower than outside. Factors which contribute to maintaining resting potential: i. Sodium-potassium pumps in axon membrane: Pumps move sodium ions out of axon and potassium ions into axon. They use energy from hydrolysis of ATP to continue moving the ions against their concentration gradient. ii. Many large, negatively charged molecules (anions) inside axon: This attracts potassium ions, reducing chance of them diffusing out of axon. iii. Impermeability of axon membrane to ions: Sodium ions cannot diffuse through axon membrane when neurone is at rest. iv. Closure of voltage-gated proteins (required for action potential in axon membrane): Stops sodium and potassium ions diffusing through axon membrane. Q9. Describe and explain the transmission of an action potential in a myelinated neurone. When an action potential is stimulated by a receptor cell in a neurone, the following steps occur:  Sodium channel proteins in axon membrane open  Sodium ions pass into axon down electrochemical gradient (there is greater conc of sodium ions outside axon than inside). Inside axon is negatively charged, attracting positively charged sodium ions.  This reduces the potential difference across axon membrane as inside of the axon becomes less negative – depolarisation  This triggers voltage-gated sodium channels to open, allowing more sodium ions to enter & cause more depolarisation  If potential difference reaches above -50mV (threshold value), many more channels open and many more sodium ions enter causing inside of axon to reach potential of about +30mV  An action potential is generated Q10. Repolarisation & refractory period  Very shortly (about 1 ms) after an action potential is generated, all sodium ion voltage gated channel proteins in this section close, stopping any further sodium ions diffusing into axon  Potassium ion voltage-gated channel proteins in this section open, allowing diffusion of potassium ions out of axon down the concentration gradient  This returns the potential difference to normal (about -70mV) – repolarisation  There is a short period of hyperpolarisation , which is when the potential difference across this section of axon membrane briefly becomes more negative than normal resting potential  Potassium ion voltage-gated channel proteins then close & sodium ion channel proteins in this section become responsive to depolarisation again  Until this occurs, this section of axon membrane is in a period of recovery & is unresponsive  This is known as refractory period Q11. Explain what is meant by saltatory conduction and describe its effect on the transmission of a nerve impulse. Saltatory conduction is the movement of an action potential along a myelinated axon, in which action potential ‘jumps’ from one node of Ranvier to the next. Local circuits set up between the nodes. This is because sodium and potassium ions cannot flow through myelin sheath, so it not possible for depolarisation or action potentials to occur in parts of axon which are surrounded by myelin sheath. Action potentials can only occur at the node of Ranvier, where all the channel proteins and pump proteins are concentrated. Q12. Describe the importance of the refractory period in the transmission of action potentials.  Ensure that action potentials are discrete events, stopping them from merging into one another  Ensures that new action potentials are generated ahead rather than behind original action potential as the region behind is ‘recovering’ from action potential that just occurred  Ensures impulse travels in one direction only  Ensure there is a minimum time between action potentials occurring at any one place along a neurone  Length of refractory period is key in determining the maximum frequency at which impulses can be transmitted Q13. Explain how the myelin sheath increases the speed of conduction of nerve impulses. In sections of axon that are surrounded by a myelin sheath, depolarisation cannot occur, as myelin sheath stops diffusion of sodium ions and potassium ions. Action potential can only occur at node of Ranvier (uninsulated sections). Local circuits of current that trigger depolarisation in the next section of axon membrane exist between nodes of Ranvier. This means action potential jumps from one node to the next, hence increasing the speed. Q14. Describe how a nerve impulse crosses a cholinergic synapse.  Arrival of an action potential at pre-synaptic membrane causes depolarisation of membrane  This stimulates voltage-gated calcium ion channel proteins to open  Calcium ions diffuse down an electrochemical gradient from tissue fluid surrounding synapse (high conc of calcium ions) into cytoplasm of pre-synaptic neurone (low conc of calcium ions)  This stimulates the acetylcholine (ACh) containing vesicles to fuse with pre-synaptic membrane, releasing ACh molecules into synaptic cleft  The ACh molecules diffuse across synaptic cleft & temporarily bind to receptor proteins in post-synaptic membrane  This causes conformational change in receptor proteins, which then open, allowing sodium ions to diffuse down an electrochemical gradient into cytoplasm of post-synaptic neurone  Sodium ions cause depolarisation of post-synaptic membrane, re-starting the impulse (that can now continue down axon of post-synaptic neurone)  To prevent sodium ion channel proteins staying permanently open & to stop permanent depolarisation of post-synaptic membrane, the ACh molecules are broken down & recycled  Enzyme acetylcholinesterase catalyses hydrolysis of ACh molecules into acetate and choline  Choline is absorbed back into pre-synaptic membrane & reacts with acetyl coenzyme a to form ACh, which is then packaged into pre-synaptic vesicles ready to be used when another action potential arrives Q15. Describe the structure of striated muscle. Striated muscle is made of muscle fibre. Muscle fibre is highly specialised cell-like unit. Cell surface membrane of muscle fibre is called sarcolemma, cytoplasm is called sarcoplasm, and endoplasmic reticulum is called sarcoplasmic reticulum (SR). Membrane of SR contains protein pump that transport calcium ions into lumen of SR. Sarcolemma has many deep tube-like projections that fold in from its outer surface. These are known as transverse system tubules or T-tubules and they run very close to SR. Sarcoplasm contains many mitochondria & myofibrils. Mitochondria to carry out aerobic respiration to generate ATP required for muscle contraction. Myofibrils are bundles of action and myosin filaments, which slide past each other during muscle contraction. Q16. Myofibrils Myofibrils are located in sarcoplasm. Each myofibril is made of two types of protein filaments: thick filaments made of myosin and thin filaments made of actin. Part of myofibril Description H band Only thick filaments present I band Only thin filaments present A band Contains areas where only thin filaments are present & areas where myosin & actin filaments overlap M line Attachment for myosin filaments Z line Attachment for actin filaments sarcomere Section of myofibril between two Z lines Q17. Describe the main structural features of thick filaments and thin filaments in the sarcomere. Thick filaments are made of myosin molecules. These are fibrous protein molecules with globular head. Fibrous part of myosin molecule anchors molecule into thick filament. Many myosin moelcues lie next to each other with their globular heads all pointing away from M line. Thin filaments are made of actin molecules. These are globular protein molecules. Many actin molecules link together, forming a chain. Two actin chains twist together to form one thin filament. A fibrous protein known as tropomyosin is twisted around two actin chains. Another protein known as troponin is attached to actin chains at regular intervals. Q18. Stimulating contraction in striated muscles.  Striated muscles contracts when it receives an impulse from a motor neurone via neuromuscular junction   When an impulse travelling along axon of a motor neurone arrives at the pre-synaptic membrane, action potential causes calcium ions to diffuse into neurone  This stimulates vesicles containing neurotransmitter acetylcholine (ACh) to fuse with pre- synaptic membrane  ACh that is released diffuses across neuromuscular junction & binds to receptor proteins on the sarcolemma (surface membrane of muscle fibre cell)  This stimulates ion channels in sarcolemma to open, allowing more sodium ions to diffuse in  This depolarises the sarcolemma, generating an action potential that passes down T-tubules towards the centre of muscle fibre  These action potentials cause voltage-gated calcium ion channel proteins in the membrane of sarcoplasmic reticulum to open  Calcium ions diffuse out of sarcoplasmic reticulum & into sarcoplasm surrounding myofibrils  Calcium ions bind to troponin molecules, stimulating them to change shape  This causes troponin and tropomyosin proteins to change position on thin (actin) filaments  Myosin-binding sites are exposed on actin molecules  Process of muscle contraction can now begin Q19. Describe how the response of the sarcoplasmic reticulum to the arrival of an action potential leads to the contraction of striated muscle (SLIDING FILAMENT MODEL).  Action potential arrives at neuromuscular junction  Calcium ions are released from sarcoplasmic reticulum  Calcium ions bind to troponin molecules, stimulating them to change shape  This causes troponin and tropomyosin proteins to change position on actin filament  Myosin binding sites are exposed to actin molecules  Globular heads of myosin molecules bind with these sites, forming cross-bridges between the two types of filament  Myosin heads move & pull actin filaments towards the centre of sarcomere causing muscle to contract a very small distance  ATP hydrolysis occurs at myosin heads, providing energy required for myosin heads to release the actin filaments  Myosin heads move back to original position & bind to new binding sites on actin filaments, closer to Z disc  Myosin heads move again, pulling actin filaments even closer the centre of sarcomere, causing sarcomere to shorten once more & pulling Z disc closer together  Myosin heads hydrolysis ATP once more in order to detach again  As long as troponin and tropomyosin are not blocking myosin-bind sites & the muscle has supply of ATP, this process repeats until muscle is fully contracted Q20. Electrical communication in Venus flytrap  The Venus flytrap is a carnivorous plant that gets its supply of nitrogen compounds by trapping and digesting small animals (mainly insects)  The specialised leaf is divided into two lobes either side of a midrib  The inside of the lobes is red and has nectar-secreting glands on the edges to attract insects  Each lobe has three stiff sensory hairs that respond to being touched  If an insect (eg. a fly) touches one of these hairs with enough force, action potentials are stimulated, which then travel very fast across the leaf  These action potentials cause the two lobes to fold together along the midrib, capturing the insect  If one of the sensory hairs is touched with enough force, calcium ion channels in cells at the base of the hair are activated  When these channels open, calcium ions flow in and generate a receptor potential  If two of the sensory hairs are stimulated within a period of about 30 seconds, or one hair is stimulated twice within this period, action potentials will travel across the trap and cause it to close  When the trap is open the lobes of the leaf are convex in shape but when the trap is triggered, the lobes quickly become concave, bending downwards and causing the trap to shut  Sealing the trap requires ongoing activation of the sensory hairs – the prey trapped inside provides this ongoing stimulation, generating further action potentials  Further stimulation of the sensory hairs stimulate calcium ions to enter gland cells where they stimulate the exocytosis of vesicles containing digestive enzymes  The trap then stays shut for up to a week to allow the prey to be digested and the nutrients from it to be absorbed by the plant Q21. Role of auxin in cell elongation growth  Auxin molecules bind to a receptor protein on the cell surface membrane  Auxin stimulates ATPase proton pumps to pump hydrogen ions from the cytoplasm into the cell wall (across the cell surface membrane)  This acidifies the cell wall (lowers the pH of the cell wall)  This activates proteins known as expansins, which loosen the bonds between cellulose microfibrils  At the same time, potassium ion channels are stimulated to open  This leads to an increase in potassium ion concentration in the cytoplasm, which decreases the water potential of the cytoplasm  This causes the cell to absorb water by osmosis (water enters the cell through aquaporins)  This increases the internal pressure of the cell, causing the cell wall to stretch (made possible by expansin proteins)  The cell elongates Q22. Role of gibberellin in germination of barley  The barley seed starts to absorb water to begin the process of germination  This stimulates the embryo to produce gibberellins  Gibberellin molecules diffuse into the aleurone layer and stimulate the cells there to synthesise amylase  In barley seeds, gibberellin does this by regulating genes involved in the synthesis of amylase, causing an increase in the transcription of mRNA coding for amylase  The amylase hydrolyses starch molecules in the endosperm, producing soluble maltose molecules  The maltose is converted to glucose and transported to the embryo  This glucose can be respired by the embryo, providing the embryo with the energy needed for it to grow CHAPTER 5 : INHERITANCE Q1. What are haploid and diploid cells?  A diploid cell is a cell that contains two complete sets of chromosomes (2n). These chromosomes contain the DNA necessary for protein synthesis and cell function  Haploid cells contain one complete set of chromosomes (n). In other words they have half the number of chromosomes compared to diploid cells o These haploid cells are called gametes and they are involved in sexual reproduction o For humans they are the female egg and the male sperm Q2. What is the need for reduction division during meiosis. During fertilization the nuclei of gametes fuse together to form the nucleus of the zygote. Both gametes must contain the correct number of chromosomes in order for the zygote to be viable. If a zygote has too many or too few chromosomes it may not survive. For a diploid zygote this means that the gametes must be haploid (n + n = 2n). Meiosis produces haploid gametes during sexual reproduction. The first cell division of meiosis is a reduction division. This is a nuclear division that reduces the chromosome number of a cell. The reduction in chromosome number during meiosis ensures the gametes formed are haploid. Q3. Describe the stages involved in meiosis. Meiosis I Prophase I  DNA condenses and becomes visible as chromosomes  DNA replication has already occurred so each chromosome consists of two sister chromatids joined together by a centromere  The chromosomes are arranged side by side in homologous pairs A pair of homologous chromosomes is called a bivalent  As the homologous chromosomes are very close together the crossing over of non-sister chromatids may occur. The point at which the crossing over occurs is called the chiasma  In this stage centrioles migrate to opposite poles and the spindle is formed  The nuclear envelope breaks down and the nucleolus disintegrates Metaphase I  The bivalents line up along the equator of the spindle, with the spindle fibres attached to the centromeres Anaphase I  The homologous pairs of chromosomes are separated as microtubules pull whole chromosomes to opposite ends of the spindle  The centromeres do not divide Telophase I  The chromosomes arrive at opposite poles  Spindle fibres start to break down  Nuclear envelopes form around the two groups of chromosomes and nucleoli reform  Some plant cells go straight into meiosis II without reformation of the nucleus in telophase I Cytokinesis   This is when the division of the cytoplasm occurs  Cell organelles also get distributed between the two developing cells  In animal cells: the cell surface membrane pinches inwards creating a cleavage furrow in the middle of the cell which contracts, dividing the cytoplasm in half  In plant cells, vesicles from the Golgi apparatus gather along the equator of the spindle (the cell plate). The vesicles merge with each other to form the new cell surface membrane and also secrete a layer of calcium pectate which becomes the middle lamella. Layers of cellulose are laid upon the middle lamella to form the primary and secondary walls of the cell  The end product of cytokinesis in meiosis I: two haploid cells  These cells are haploid as they contain half the number of centromeres Meiosis II Prophase II  The nuclear envelope breaks down and chromosomes condense  A spindle forms at a right angle to the old one Metaphase II  Chromosomes line up in a single file along the equator of the spindle Anaphase II  Centromeres divide and individual chromatids are pulled to opposite poles  This creates four groups of chromosomes that have half the number of chromosomes compared to the original parent cell Telophase II  Nuclear membranes form around each group of chromosomes Cytokinesis  Cytoplasm divides as new cell surface membranes are formed creating four haploid cells  The cells still contain the same number of centromeres as they did at the start of meiosis I but they now only have half the number of chromosomes (previously chromatids) Q4. What are the sources of genetic variation. 1. Crossing over: Crossing over is the process by which non-sister chromatids exchange alleles. Process:  During meiosis I homologous chromosomes pair up and are in very close proximity to each other  The non-sister chromatids can cross over and get entangled  These crossing points are called chiasmata  The entanglement places stress on the DNA molecules  As a result of this a section of chromatid from one chromosome may break and rejoin with the chromatid from the other chromosome This swapping of alleles is significant as it can result in a new combination of alleles on the two chromosomes. There is usually at least one, if not more, chiasmata present in each bivalent during meiosis. Crossing over is more likely to occur further down the chromosome away from the centromere. 2. Independent assortment:  Independent assortment is the production of different combinations of alleles in daughter cells due to the random alignment of homologous pairs along the equator of the spindle during metaphase I  The different combinations of chromosomes in daughter cells increases genetic variation between gametes  In prophase I homologous chromosomes pair up and in metaphase I they are pulled towards the equator of the spindle  Each pair can be arranged with either chromosome on top, this is completely random  The orientation of one homologous pair is independent / unaffected by the orientation of any other pair  The homologous chromosomes are then separated and pulled apart to different poles  The combination of alleles that end up in each daughter cell depends on how the pairs of homologous chromosomes were lined up 3. Random fusion of gametes:  During fertilization any male gamete can fuse with any female gamete to form a zygote  This random fusion of gametes at fertilization creates genetic variation between zygotes as each will have a unique combination of alleles Q5. What is chi-squared test and why do we use it? A statistical test called the chi-squared test determines whether there is a significant difference between the observed and expected results in an experiment. The chi-squared test is completed when the data is categorical (data that can be grouped). STEPS: 1. Obtain the expected and observed results for the experiment 2. Calculate the difference between each set of results 3. Square each difference (as it is irrelevant whether the difference is positive or negative) 4. Divide each squared difference by the expected value and get a sum of these answers to obtain the chi-squared value ANALYSIS:  To work out what the chi-squared value means, a table that relates chi-squared values to probabilities is used  If the chi-squared value represents a larger probability than the critical probability then it can be stated that the differences between the expected and observed results are due to chance  If it represents a smaller probability than the critical probability then the differences in results are significant and something else may be causing the differences  To determine the critical probability biologists generally use a probability of 0.05 (they allow that chance will cause five out of every 100 experiments to be different)   The number of comparisons made must also be taken into account when determining the critical probability. This is known as the degrees of freedom. Calculated by subtracting one from the number of classes. Q6. TYR gene and albinism.  The amino acid tyrosine is converted to DOPA by the enzyme tyrosinase  DOPA is converted to dopaquinone again by the enzyme tyrosinase  Dopaquinone is converted to melanin tyrosine → DOPA → dopaquinone → melanin  A gene called TYR located on chromosome 11 codes for the enzyme tyrosinase  There is a recessive allele for the gene TYR that causes a lack of enzyme tyrosinase or the presence of inactive tyrosinase  Without the tyrosinase enzyme tyrosine can not be converted into melanin Q7. HBB gene and sickle cell anaemia.  β-globin is a polypeptide found in haemoglobin that is coded for by the gene HBB which is found on chromosome 11  There is an abnormal allele for the gene HBB which produces a slightly different amino acid sequence to the normal allele  The change of a single base in the DNA of the abnormal allele results in an amino acid substitution (the base sequence CTC is replaced by CAC)  This change in amino acid sequence (the amino acid Glu is replaced with Val) results in an abnormal β-globin polypeptide  The abnormal β-globin in haemoglobin affects the structure and shape of the red blood cells  They are pulled into a half moon shape  They are unable to transport oxygen around the body  They stick to each other and clump together blocking capillaries Q8. F8 gene and haemophilia.  Factor VIII is a coagulating agent that plays an essential role in blood clotting  The gene F8 codes for the Factor VIII protein  There are abnormal alleles of the F8 gene that result in: o Production of abnormal forms of factor VIII o Less production of normal factor VIII o No production of factor VIII  A lack of normal factor VIII prevents normal blood clotting and causes the condition haemophilia  The F8 gene is located on the X chromosome  o This means F8 is a sex-linked gene o Haemophilia is a sex-linked condition o If males have an abnormal allele they will have the condition as they have only one copy of the gene o Females can be heterozygous for the F8 gene and not suffer from the condition but act as a carrier Q9. HTT gene and Huntington’s disease.  Huntington’s disease is a genetic condition that develops as a person ages  Usually a person with the disease will not show symptoms until they are 30 years old +  An individual with the condition experiences neurological degeneration; they lose their ability to walk, talk and think  The disease is ultimately fatal  It has been found that individuals with Huntington's disease have abnormal alleles of the HTT gene o The HTT gene codes for the protein huntingtin which is involved in neuronal development o People that have a large number (>40) of repeated CAG triplets present in the nucleotide sequence of their HTT gene suffer from the disease  The abnormal allele is dominant over the normal allele o If an individual has one abnormal allele present they will suffer from the disease Q10. The role of gibberellin in stem elongation.  The Le gene dictates the height of some plants  It has two alleles: Le and le o The dominant allele Le produces tall plants when present o The recessive allele le produces shorter plants when present (in a homozygous individual)  The gene regulates the production of an enzyme that is involved in a pathway that forms active gibberellin GA1  Active gibberellin is a hormone that helps plants grow by stimulating cell division and elongation in the stem  The recessive allele le results in non-functional enzyme o It is only one nucleotide different to the dominant allele o This causes a single amino acid substitution (threonine -> alanine) in the primary structure of the enzyme o This change in primary structure occurs at the active site of the enzyme, making it non- functional  Without this enzyme no active gibberellin is formed and plants are unable to grow tall  Plants that are homozygous for the recessive allele le are dwarves Q11. What are structural and regulatory genes. A structural gene codes for a protein that has a function within a cell. A regulatory gene codes for a protein that helps to control the expression of another gene Q12. What are inducible enzymes and repressible enzymes.  Inducible enzymes are only synthesized when their substrate is present. The presence of the substrate induces the synthesis of of the enzyme by causing the transcription of the gene for the enzyme to start  Repressible enzymes are synthesized as normal until a repressor protein binds to an operator. The presence of the repressor protein represses the synthesis of the enzyme by causing the transcription of the gene for the enzyme to stop Q13. Structure of the lac operon. The components of the lac operon are found in the following order:  Promoter for structural genes  Operator  Structural gene lacZ that codes for lactase  Structural gene lacY that codes for permease (allows lactose into the cell)  Structural gene lacA that codes for transacetylase Located to the left (upstream) of the lac operon on the bacterium's DNA there is also the:  Promoter for regulatory gene  Regulatory gene lacI that codes for the lac repressor protein Q14. Presence of lactose in Lac Operon. There is an uptake of lactose by the bacterium The lactose binds to the second binding site on the repressor protein, distorting its shape so that it cannot bind to the operator site RNA polymerase is then able to bind to the promoter region and transcription takes place The mRNA from all three structural genes is translated Enzyme lactase is produced and lactose can be broken down and used for energy by the bacterium Q15. Absence of lactose in Lac Operon. The regulatory gene is transcribed and translated to produce lac repressor protein The lac repressor protein binds to the operator region upstream of lacZ Due to the presence of the repressor protein RNA polymerase is unable to bind to the promoter region Transcription of the structural genes does not take place No lactase enzyme is synthesized Q16. How transcription factors work? Some transcription factors bind to the promoter region of a gene.This binding can either allow or prevent the transcription of the gene from taking place. The presence of a transcription factor will either increase or decrease the rate of transcription of a gene. Q17. Gene control by gibberellin. The breakdown of DELLA protein by gibberellin is necessary for the synthesis of amylase. The following components are involved:  Repressor protein DELLA   Transcription factor PIF  Promoter of amylase gene  Amylase gene  Gibberellin  Gibberellin receptor and enzyme The process occurs as follows:  DELLA protein is bound to PIF, preventing it from binding to the promoter of the amylase gene so no transcription can occur  Gibberellin binds to a gibberellin receptor and enzyme which starts the breakdown of DELLA  PIF is no longer bound to DELLA protein and so it binds to the promoter of the amylase gene  Transcription of amylase gene begins  Amylase is produced CHAPTER 6 : SELECTION & EVOLUTION Q1. Explain phenotypic variation. The observable characteristics of an organism are its phenotype. Phenotypic variation is the difference in phenotypes between organisms of the same species. Phenotypic variation is explained by genetic factors. For example, the four different blood groups observed in human populations are due to different individuals within the population having two of three possible alleles for the single ABO gene. Phenotypic variation can be explained by environmental factors. For example, clones of plants with exactly the same genetic information (DNA) will grow to different heights when grown in different environmental conditions. Phenotypic variation can also be explained by a combination of genetic and environmental factors. For example, the recessive allele that causes sickle cell anaemia has a high frequency in populations where malaria is prevalent due to heterozygous individuals being resistant to malaria. The complete phenotype is determined by expression of genotype and the interaction of the environment on this. Q2. Sources of genetic variation.  Independent assortment of homologous chromosomes during metaphase I  Crossing over of non-sister chromatids during prophase I  Random fusion of gametes during fertilization  Mutation - random change in DNA base sequence resulting in generation of new allele Q3. Environmental factors to be considered. The environment that an organism lives in can also have an impact on its phenotype. Changes in the factors including temperature, pH, oxygen levels, nutrient levels above can affect how organisms grow and develop. Variation in phenotype caused solely by environmental pressures or factors cannot be inherited by an organism's offspring. Q4. Explain why variation in phenotype due to genetics is inherited but variation due to environmental factors is not. Variation in the phenotype due to genetics affects the DNA of the gametes. When those gametes fuse to form a zygote, the genes/allele are passed on to the offspring. So the variation is inherited by the offspring. However, the variation in phenotype due to environmental factors doesn’t affect the DNA of the gametes. Hence it is not passed on to the offspring and not inherited by them. Q5. Explain what is meant by continuous and discontinuous variation. Continuous variation occurs when there are quantitative differences in the phenotypes of individuals within a population for particular characteristics. Quantitative differences do not fall into discrete categories. No distinct classes or categories exist. Instead a range of values exist between two extremes within which the population will fall. Example: height, weight of group of individuals. Discontinuous variation occurs when there are qualitative difference in phenotype of individuals within a population. They fall into discrete and distinguishable categories, with no intermediates. Example: four possible ABO blood groups in human (there are no intermediates). Q6. Explain the genetic basis of continuous variation and discontinuous variation. Discontinuous variation occurs solely due to genetic factors. The environment has no direct effect. Different genes have different effects on the phenotype and different alleles at a single gene locus have a large effect on the phenotype in discontinuous variation. Continuous variation is caused by an interaction between the genetics and the environment. Different alleles at a single gene locus have a small effect on the phenotype. Different genes can have the same effect on the phenotype and these add together to have an additive effect. If a large number of genes have a combined effect on phenotype, they are known as polygenes. Q7. What is a t-test and how do we carry it out. T-test can be used to compare the means of two set of data and determine whether they are significantly different or not. The sets of data must follow a rough normal distribution, be continuous & the standard deviations should be approximately constant. STEPS: 1. Null hypothesis: there is no statistically significant difference between the means of sample 1 and sample 2 2. Calculate mean for each data set 3. Calculate the standard deviation for each set of data, s1 = standard deviation of sample 1 and s2 = standard deviation of sample 2: 4. Square the standard deviation and divide by n (the number of observation) in each sample, for both sample: 5. Add the values from step 4 together and take the square root: 6. Divide the difference between the two means (from step 2) with the value calculate in step 5 to get the t value: 7. Calculate the degrees of freedom (v) for the whole data set: 8. Look at a table that relates t values to the probability that the differences between data sets is due to chance to find where the t value for the degrees of freedom (v) calculated lies 9. The greater the t value calculated (for any degree of freedom), the lower the probability of chance causing any significant difference between the two sample means   Identify where the t value calculated lies with respect to the confidence levels provided  If the t value is greater than the critical value (obtained from the table at the critical probability of 0.05) then any difference between the two data sets is less likely to be due to chance, so the null hypothesis can be rejected  If the t value is less than the critical value given at a confidence of 5% / the probability that any difference is down to chance is above 0.05; then an assumption can be made that the differences between the means of the two sets of data are not significant and the null hypothesis is accepted Q8. What is null hypothesis and how do we write it? Null hypothesis is basically a statement of what we would expect if there is no significant difference between the two means, and that any differences are due to chance. Framing a null hypothesis: there is no significant difference between the mean of data A and data B. Q9. Environmental factors which limit the population size. Environmental factors limit population sizes by reducing the rate of population growth whenever a population reaches a certain size. Environmental factors can be biotic or abiotic. Biotic factors involve other living organisms. This includes things like predation, competition for resources and disease. Abiotic factors involve the non-living parts of an environment. Examples of abiotic factors include light availability, water supply and soil pH. When biotic and abiotic factors come into play not all individuals within a population will survive. Q10. What is meant by natural selection? Natural selection is the process by which individuals with a fitter phenotype are more likely to survive and pass on their alleles to their offspring so that the advantageous alleles increase in frequency over time and generations. Q11. Explaining the concept of natural selection.  Variation exists within a species population  This means that some individuals within the population possess different phenotypes (due to genetic variation in the alleles they possess)  Environmental factors affect the chance of survival of an organism; they, therefore, act as a selection pressure  Selection pressures increase the chance of individuals with a specific phenotype surviving and reproducing over others  The individuals with the favoured phenotypes are described as having a higher fitness o The fitness of an organism is defined as its ability to survive and pass on its alleles to offspring o Organisms with higher fitness posses adaptations that make them better suited to their environment  When selection pressures act over several generations of a species they have an effect on the frequency of alleles in a population through natural selection Q12. Case study: Natural Selection in rabbits   Variation in their fur colour exists within rabbit populations  At a single gene locus, normal brown fur is produced by a dominant allele whereas white fur is produced by a recessive allele in a homozygous individual  Rabbits have natural predators like foxes which act as a selection pressure  Rabbits with a white coat do not camouflage as well as rabbits with brown fur, meaning predators are more likely to see white rabbits when hunting  As a result, rabbits with white fur are less likely to survive than rabbits with brown fur  The rabbits with brown fur therefore have a selection advantage, so they are more likely to survive to reproductive age and be able to pass on their alleles to their offspring  Over many generations, the frequency of alleles for brown fur will increase and the frequency of alleles for white fur will decrease Q13. Three types of natural selection. directional 1. Stabilising selection is natural selection that keeps allele frequencies relatively constant over generations. This means things stay as they are unless there is a change in the environment. 2. Directional selection is natural selection that produces stabilising a gradual change in allele frequencies over several generations. This usually happens when there is a change in environment / selection pressures or a new allele has appeared in the population that is advantageous. 3. Disruptive selection is natural selection that maintains high disruptive frequencies of two different set of alleles. Individuals with intermediate phenotypes or alleles are selected against. Disruptive selection causes polymorphism; continued existence of two or more distinct phenotypes in the species. This can occur in an environment that shows variation. Q14. Explain genetic drift. When a population is very small chance can affect which alleles get passed on to the next generation. Meiosis results in haploid gametes, meaning that a fertilisation event only passes on half of the alleles of an individual; the half that gets passed on is the result of random fertilisation, and the other half of the alleles may be lost to the next generation. Over time some alleles can be lost or passed on purely by chance; this is genetic drift. Genetic drift is more likely to affect allele frequencies in a small population. Example: In a small population of five plants growing near a playground with a rubber floor, three of the plants have blue flowers and two of the plants have pink flowers. By chance most of the seeds from the pink flowered plants end up on the rubber floor of the playground while all the seeds from the blue flowered plants land on fertile soil where they are able to germinate and grow. Note that the seeds from the pink flower do not fall on the impermeable surface because of any disadvantageous allele in the plant's genome, but purely by chance, e.g. because of a gust of wind or a passing animal. If this happens by chance over several generations the allele for the pink flowers may be lost from this population. Q15. What is founder effect? The founder effect occurs when a small number of individuals from a large parent population start a new population. The founder effect can come about as the result of chance. As the new population is made up of only a few individuals from the original population only some of the total alleles from the parent population will be present. In other words, not all of the gene pool is present in the smaller population. Because the population that results from the founder effect is very small it is more susceptible to the effects of genetic drift. Q16. Case Study: Founder effect in Lizards. Anole lizards inhabit most Caribbean Islands and they can travel from one island to another via floating debris or vegetation. A small number of lizards may be separated from the main population on a larger island and carried away to a smaller island by a chance event such as a large ocean wave or a storm. The lizards arriving at a new island may only carry a small selection of alleles between them, with many more alleles present in the lizard population on the original island. E.g. the lizards on the original island could display a range of scale colors from white to yellow and the two individual lizards that arrived on the island may have white scales. This means that the whole population that grows on that island might only have individuals with white scales. In comparison the original island population has a mixture of white and yellow scaled individuals. If the yellow allele were recessive and present as a single copy in the original two lizards that arrived on the island, the chance of it being lost as a result of genetic drift is increased due to the small size of the gene pool. Q17. Explain the Bottleneck effect. The bottleneck effect is similar to the Founder effect. It occurs when a previously large population suffers a dramatic fall in numbers. A major environmental event can greatly reduce the number of individuals in a population which in turn reduces the genetic diversity in the population as alleles are lost. The surviving individuals end up breeding and reproducing with close relatives. This leads to an increase in homozygosity among individuals. Q18. Case Study: Bottleneck effect in Cheetahs. Roughly 10,000 years ago there was a large and genetically diverse cheetah population. Most of the population was suddenly killed off when the climate changed drastically at the end of the Ice Age. As a result the surviving cheetahs were isolated in small populations and lots of inbreeding occurred. This meant that the cheetah population today has a lack of genetic variation. This is problematic for conservation as genetic variation within a species increases the likelihood that the species is able to respond in the event of any environmental changes. Q19. Antibiotic resistance in bacteria due to natural selection.   Antibiotics are either described as being bactericidal (they kill) or bacteriostatic (they inhibit growth processes), they target prokaryotic features but can affect both pathogenic and mutualistic bacteria living on or in the body  However, like in all species, there exists genetic diversity within populations, and the same applies to disease-causing bacteria  Individual bacterial cells may possess alleles that confer resistance to the effects of the antibiotic. These alleles are generated through random mutation and are not caused by antibiotic use, but antibiotic use exerts selection pressures that can result in the increase in their frequency.  Bacteria have a single loop of DNA with only one copy of each gene so when a new allele arises it is immediately displayed in the phenotype  When an antibiotic is present: o Individuals with the allele for antibiotic resistance have a massive selective advantage so they are more likely to survive, reproduce and pass their genome (including resistance alleles) on o Those without alleles are less likely to survive and reproduce o Over several generations, the entire population of bacteria may be antibiotic- resistant By using antibiotics frequently, humans exert a selective pressure on the bacteria, which supports the evolution of antibiotic resistance. Q20. What is Hardy-Weinberg principle and why do we use it? The Hardy-Weinberg principle states that if certain conditions are met then the allele frequencies of a gene within a population will not change from one generation to the next. The Hardy- Weinberg equation allows for the calculation of allele and genotype frequencies within populations. It also allows for predictions to be made about how these frequencies will change in future generations. Q21. What are the 7 conditions that must be met for Hardy-Weinberg principle to be applied correctly to a population? 1. Organisms are diploid 2. Organisms reproduce by sexual reproduction only 3. There is no overlap between generations, i.e. parents do not mate with offspring 4. Random mating 5. The population is large 6. There is no migration, mutation, or selection a. This would mean no individuals entering the population (immigration) or leaving (emigration) b. Selection refers to both natural and artificial selection 7. Allele frequencies are equal in both sexes Q22. Using the Hardy-Weinberg equation. Three possible genotype: homozygous recessive, heterozygous, homozygous dominant. When using the Hardy-Weinberg equation frequencies are represented as proportions of the population; a proportion is a number out of 1. Frequency of alleles can be represented; this is the proportion of all of the alleles in a population that are of a particular form o The letter p represents the frequency of the dominant allele o The letter q represents the frequency of the recessive allele o As there are only two alleles at a single gene locus for a phenotypic trait in the population: p+q=1 Frequency of genotypes can also be represented; this is the proportion of all of the individuals with a particular genotype o The chance of an individual being homozygous dominant is p2  The offspring would inherit dominant alleles from both parents so p x p = p2 o The chance of an individual being heterozygous is 2pq  Offspring could inherit a dominant allele from the father and a recessive allele from the mother (p x q) or offspring could inherit a dominant allele from the mother and a recessive allele from the father (p x q) so 2pq o The chance of an individual being homozygous recessive is q2  The offspring would inherit recessive alleles from both parents so q x q = q2 o As these are all the possible genotypes of individuals in the population the following equation can be constructed: p2 + q2 + 2pq = 1 Q23. What is artificial selection?  Artificial selection is the process by which humans choose organisms with desirable traits and selectively breed them together to enhance the expression of these desirable traits over time and many generations  This practice is also known as selective breeding  Knowledge of the alleles that contribute to the expression of the desired traits are not required as individuals are selected by their phenotypes, and not their genotypes Q24. Principles of selective breeding. 1. The population shows phenotypic variation - there are individuals with different phenotypes / traits 2. Breeder selects an individual with the desired phenotype 3. Another individual with the desired phenotype is selected. The two selected individuals should not be closely related to each other 4. The two selected individuals are bred together 5. The offspring produced reach maturity and are then tested for the desirable trait. Those that display the desired phenotype to the greatest degree are selected for further breeding 6. The process continues for many generations: the best individuals from the offspring are chosen for breeding until all offspring display the desirable trait Q25. Examples of selective breeding.  The introduction of disease resistance to varieties of wheat and rice:  Wheat crops can be badly affected by fungal diseases: Fusarium is a fungus that causes “head blight” in wheat plants  Fungal diseases are highly problematic for farmers as they destroy the wheat plant and reduce crop yield  By using selective breeding to introduce a fungus-resistant allele from another species of wheat, the hybrid wheat plants are not susceptible to infection, and so yield increases  Rice is another crop that has been subject to large amounts of selective breeding  Rice plants are prone to different bacterial and fungal diseases o Examples include “bacterial blight” and “rice blast” caused by the Magnaporthe fungus  These diseases all reduce the yield of the crop as they damage infected plants  Inbreeding and hybridisation to produce vigorous, uniform varieties of maize:  Maize plants have been heavily inbred (bred with plants with similar genotypes to their own)  This has resulted in small and weaker maize plants that have less vigour  This is inbreeding depression which: o Increases the chance of harmful recessive alleles combining in an individual and being expressed in the phenotype o Increases homozygosity in individuals (paired alleles at loci are identical) o Leads to decreased growth and survivability  A farmer can prevent inbreeding depression by outbreeding o This involves breeding individuals that are not closely related o Outbreeding produces taller and healthier maize plants o It decreases the chance of harmful recessive alleles combining in an individual and being expressed in the phenotype o Increases heterozygosity (paired alleles at loci are different) o Leads to increased growth and survivability (known as hybrid vigour) o Crops of these plants have a greater yield  Uniformity is important when growing a crop: o If outbreeding is carried out completely randomly, it can produce too much variation between plants within one field o A farmer needs the plants to ripen at the same time and be of a similar height; the more variation there is, the less likely this is  In order to achieve heterozygosity and uniformity, farmers buy sets of homozygous seeds from specialised companies and cross them to produce an F1 generation  Different hybrids of maize are constantly being created and tested for desirables traits such as: resistance to pests / disease, higher yields and good growth in poor conditions  Improving the milk yield of diary cattle:  Over many years and generations farmers have selected female cows that have the highest milk yield and crossed them with male bulls related to high yield females  Over time this selective breeding has resulted in cows with greater milk yields, which has been of great economical benefit to farmers   In cows it has been observed that selectively bred individuals are much more prone to ailments such as mastitis (inflammation of the udder), milk fever and lameness compared to those that were allowed to breed at random Q26. Explain theory of evolution. A species can be defined as a group of organisms that are able to interbreed and produce fertile offspring. Members of one species are reproductively isolated from members of another species. Individuals of the same species have similar behavioural, morphological (structural) and physiological (metabolic) features. When the gene pool within a species population changes sufficiently over time (due to genetic drift/natural selection/founder effect), the characteristics of the species will also change. The change can become so great that a new species forms. This is evolution. Evolution is the formation of new species from pre-existing species over time, as a result of changes to gene pools from generation to generation For a population to have evolved into a separate species it must be genetically and reproductively isolated from the pre-existing species population. Reproductive isolation can occur due to mutations that lead to the incompatibility of gametes or sex organs, or differences in breeding behaviour. When two populations are reproductively isolated, they can also be said to be genetically isolated from each other, meaning that they do not exchange genes with each other in the production of offspring. Changes in the allele frequencies of isolated populations are not shared so they evolve independently of each other; this can lead to the formation of two groups that are no longer successfully able to interbreed and that are said to be separate species. The formation of new species in this way is known as speciation. The evolution of a new species can take a very long time and many generations. Q27. How DNA sequence data can show evolutionary relationships between species. DNA found in the nucleus, mitochondria and chloroplasts of cells can be sequenced and used to show evolutionary relationships between species. The differences between the nucleotide sequences (DNA) of different species indicate:  The more similar the sequence the more closely related the species are  Two groups of organisms with very similar DNA will have separated into separate species more recently than two groups with less similarity in their DNA sequences Q28. Suggest why mtDNA is used instead of nuclear DNA when studying the closeness of the relationship between populations.  A zygote only contains the mitochondria of the egg and none from the sperm so only maternal mitochondrial DNA is present in a zygote.  There is no crossing over that occurs in mtDNA so the base sequence can only change by mutation. According to the molecular theory, mutations occur at a constant rate, mtDNA mutates faster (than nuclear DNA).  There are many copies of mtDNA per cell.  mtDNAis smaller and has fewer genes.  mtDNA is not associated with histones.  mtDNA analysis is quicker compared to nuclear DNA. Q29. Explain what is meant by allopatric speciation. Allopatric speciation occurs as a result of geographical isolation. A species population splits into one or more groups which then become separated from each other by geographical barriers. The barrier could be natural like a body of water, or a mountain range. It can also be man-made (like a motorway). This separation creates two populations of the same species who are isolated from each other, and as a result, no genetic exchange can occur between them. If there is sufficient selection pressure or genetic drift acting to change the gene pools within both populations then eventually these populations will diverge and form separate species. The changes in the alleles/genes of each population will affect the phenotypes present in both populations. Over time, the two populations may begin to differ physiologically, behaviourally and morphologically (structurally). Q30. Explain what is meant by sympatric speciation. Sympatric speciation takes place with no geographical barrier. A group of the same species could be living in the same place but in order for speciation to take place there must exist two populations within that group and no gene flow occurs between them. Something has to happen that splits or separates the population: o Ecological separation: Populations are separated because they live in different environments within the same area  For example, soil pH can differ greatly in different areas. Soil pH has a major effect on plant growth and flowering o Behavioural separation: Populations are separated because they have different behaviours  For example differences in feeding, communication or social behaviour CHAPTER 7 : CLASSIFICATION, BIODIVERSITY & CONSERVATION Q1. Describe the three specis concept. Biological species concept: a group of organsims with similar morphology and physiology, which can breed together to produce fertile offspring and are reproductively isolated from each other. Morphological species: a group of organisms that share many physical features that distinguish them from other species. Ecological species: a population of individuals of the same species living in the same area at the same time. Q2. Binomial nomenclature. The first part of the name is the genus that the species belongs to; this is a group of very similar organisms. The second part of the name is specific and unique to a single group of organisms that are identified as a species (and occasionally there may be a third name). The binomial name is always italicized in writing. Q3. Name the three domains and describe their features. Archaea  Organisms within this domain are sometimes referred to as the extremophile prokaryotes, archaea were first discovered living in extreme environments, but not all archaea do  Archael cells have no nucleus (and so are prokaryotic)  They were initially classified as bacteria until several unique properties were discovered that separated them from known bacteria, including: o Unique lipids being found in the membranes of their cells o No peptidoglycan in their cell walls o Ribosomal structure (particularly that of the small subunit) are more similar to the eukaryotic ribosome than that of the bacteria  Archaea a similar size range as bacteria (and in many ways metabolism is similar between the two groups)  DNA transcription is more similar to that of eukaryotes Bacteria  These are organisms that have prokaryotic cells which contain no nucleus  They vary in size over a wide range: the smallest are bigger than the largest known-viruses and the largest are smaller that the smallest known single-celled eukaryotes  Bacterial cells divide by binary fission Eukarya  Organisms that have eukaryotic cells with nuclei and membrane-bound organelles are placed in this domain  They vary massively in size from single-celled organisms several micrometres across to large multicellular organisms many-metres in size, such as blue whales  Eukaryotic cells divide by mitosis  Eukaryotes can reproduce sexually or asexually Q4. What are the differences between archaea and bacteria. Membrane lipids:  The membrane lipids found in the cells of Archaea organisms are completely unique  They are not found in any bacterial or eukaryotic cells  The membrane lipids of Archaea consist of branched hydrocarbon chains bonded to glycerol by ether linkages  The membrane lipids of Bacteria consist of unbranched hydrocarbon chains bonded to glycerol by ester linkages Ribosomal RNA:  Both Archaea and Bacteria possess 70S ribosomes  The 70S ribosomes in Archaea possess a smaller subunit that is more similar to the subunit found in Eukaryotic ribosomes than subunits in Bacterial ribosomes  The base sequences of ribosomal RNA in Archaea show more similarity to the rRNA of Eukarya than Bacteria  The primary structure of ribosome proteins in Archaea show more similarity to the ribosome proteins in Eukarya than Bacteria Composition of cell walls:  Organisms from the Bacteria domain have cells that always possess cell walls with peptidoglycan  Organisms from the Archaea domain also have cells that always possess cell walls, however these do not contain peptidoglycan Q5. Hierarchical classification system. Similar species can be grouped in a genus Similar genuses can be grouped in a family Similar families can be grouped into an order Similar orders can be grouped into a class Similar classes can be grouped into a phylum Similar phyla can be grouped into a kingdom Similar kingdoms can be grouped into a domain Q6. Name the 4 kingdom of domain Eukarya and describe their features. Kingdom Protoctista  All Protoctista are eukaryotic, and this broad group of cellular life encompasses all eukaryotic cells that do not belong to the other three eukaryotic kingdoms  Members of this kingdom show great diversity in all aspects of life including structure, life cycle, feeding and trophic levels and well as modes of locomotion  Protoctists can exist as single-celled organisms or as a group of similar cells  A group of Protoctista known as protozoa possess cells similar to animal cells Their cells have no cell wall  Another group of Protoctista known as algae possess cells similar to plant cells Their cells have cellulose cell walls and chloroplasts Kingdom Fungi  All fungi are eukaryotic cells  The cells of fungi: o Possess non-cellulose cell walls (often made of the polysaccharides chitin and glucans  o Don’t have cilia  Fungi are heterotrophs: o They use organic compounds made by other organisms as their source of energy and molecules for metabolism o They obtain this energy and carbon by digesting dead/decaying matter extracellularly or from being parasites on living organisms  Fungi reproduce using spores that disperse onto the ground nearby  Fungi have a simple body form: o They can be unicellular (like the common baker’s yeast Saccharomyces cerevisiae o Some consist of long threads called hyphae that grow from the main fungus body (mycelium) o Larger fungi possess fruiting bodies that release large numbers of spores Kingdom Plantae  Plants are multicellular eukaryotic organisms  Plant cells: o All have cell walls composed of cellulose o Possess large (and usually permanent) vacuoles that provide structural support o Are able to differentiate into specialized cells to form tissues and organs o Possess chloroplasts that enable photosynthesis (not all plant cells have chloroplasts) o Can sometimes have flagella  They are autotrophs o This means they can synthesize their organic compounds and molecules for energy use and building biomass from inorganic compounds  Plants have complex body forms o They have branching systems above and below the ground Kingdom Animalia  Animals are also multicellular eukaryotic organisms  Animal cells: o Are able to differentiate into many different specialised cell types that can form tissues and organs o Have small temporary vacuoles (for example, lysosomes) o Have no cell walls o Sometimes have cilia  They are heterotrophs o They have a wide range of feeding mechanisms  They have a wide range of body forms: o Communication within their complex body forms takes place through a nervous system and chemical signalling Q7. Classification of viruses. Viruses are classified according to the type of nucleic acid (RNA or DNA) their genome is made from, and whether it is single-stranded or double-stranded. In viruses, DNA and RNA can be either single- stranded or double-stranded. As a result, there are four groups of viruses that exist: 1. DNA single-stranded viruses 2. DNA double-stranded viruses 3. RNA single-stranded viruses 4. RNA double-stranded viruses Q8. Define the terms ecosystem and niche. An ecosystem is a relatively self-contained community of interacting organisms and the environment they live in, and interact with. There are both living (biotic) components and non-living (abiotic)components within an ecosystem. The place where a species lives within an ecosystem is its habitat The role that species plays within an ecosystem is its niche.  It encompasses where in the environment the organism is, how it gets its energy and how it interacts with other species and its physical environment  This is how an organism fits into the ecosystem Q9. What is biodiversity and what are its different levels. Biodiversity can be thought of as a study of all the variation that exists within and between all forms of life. Biodiversity looks at the range and variety of genes, species and habitats within a particular region. Ecosystem/habitat diversity: This is the range of different ecosystems or habitats within a particular area or region. Species diversity: Species diversity looks at the number of different species in an ecosystem, and also the evenness of abundance across the different species present. The greater the number of species in an ecosystem, and the more evenly distributed the number of organisms are among each species, then the greater the species diversity. Genetic diversity: The genetic diversity within a species is the diversity of alleles and genes in the genome of species Q10. Random sampling and systematic sampling. In random sampling the positions of the sampling points are completely random or due to chance  This method is beneficial because it means there will be no bias by the person that is carrying out the sampling that may affect the results. In systematic sampling the positions of the sampling points are chosen by the person carrying out the sampling  There is a possibility that the person choosing could show bias towards or against certain areas  Individuals may deliberately place the quadrats in areas with the least species as these will be easier and quicker to count  This is unrepresentative of the whole area Q11. Frame transects  When carrying out sampling, square frames called quadrats can be used to mark off the area being sampled  Quadrats of different sizes can be used depending on what is being measured and what is most suitable in the space the samples are being made in  Quadrats must be laid randomly in the area to avoid sampling bias o This random sampling can be done by converting the sampling area into a grid format and labelling each square on the grid with a number o Then a random number generator is used to pick the sample points  Once the quadrat has been laid on the chosen sample point the abundance of all the different species present can be recorded  Species frequency is the probability that the species will be found within any quadrat in the sample area o The number of quadrats that the species was present in is divided by the total number of quadrats and then multiplied by 100  Species density indicates how many individuals of that species there are per unit area o The number of individuals counted across all quadrats is divided by the total area of all the quadrats Q12. Line and