Unit 5 - 7 - Processes

Question 1

Light Dependent Reaction - Non-Cyclic Photophosphorylation in 6 steps

Answer 1

1. Light energy is absorbed by PSII. The light energy excites electrons in chlorophyll which are released.
2. The electrons move down an electron transport chain and to a higher energy level and go to PSI
3. Light energy is used to split water into protons, electrons and oxygen
4. The energy from the excited electrons is used to actively pump protons from stroma into the thylakoid membrane which causes a concentration gradient to form
5. Protons move down the concentration gradient back into the stroma via ATP synthase in which the movement drives synthesis of ADP + Pi to produce ATP
6. Light energy is absorbed by PSI which causes electrons to raise to a higher energy level and electrons are transferred to make reduced NADP

Question 2

Light Independent Reaction - Calvin Cycle - in 3 steps

Answer 2

1) RuBP fixes to CO2 via rubisco to form 2 x GP
2) 2X GP is reduced through 2 NADPH and 2 ATP to form 2 X TP
3)2 xTP uses 5 of 6 carbons to regenerate RuBP with ATP and uses the extra carbon to form organic compounds

Question 3

Krebs Cycle - in 3 steps

Answer 3

1) Acetyl Coenzyme A combines with 4C compound to form 6C compound, and sends CoA back to the link reaction

2) 6C compound undergoes dehydrogenation and decarboxylation to form reduced NAD, CO2 and a 5C compound

3) 5C compound undergoes dehydrogenation and decarboxylation to form 2 reduced NAD, CO2, reduced FAD and substrate level phosphorylation occurs to produce ATP - 4C compound is regenerated

Question 4

Oxidative Phosphorylation - in 6 steps

Answer 4

1) Reduced coenzymes release hydrogen and are oxidised and hydrogen splits into protons and electrons

2) Electrons go down the electron transport chain releasing energy at each carrier

3) The energy from each electron carrier is used to actively pump protons into the intermembranal space

4) Protons move down their electrochemical gradient back into the matrix via ATP synthase

5) This movement helps synthesis of ADP + Pi to form ATP in a chemiosmosis reaction

6) Protons, electrons and oxygen combine to form water. Oxygen is the final electron acceptor

Question 5

The Nitrogen Cycle - Each of the 4 steps

Answer 5

1) Denitrification - Nitrates to Nitrogen Gas
- Done by denitrifying bacteria
- Happens in anaerobic condition (waterlogged soils)

2) Nitrogen Fixation - Nitrogen gas to Ammonia
- Done by nitrogen-fixing bacteria

3) Ammonification - Nitrogen compounds to Ammonia to Ammonium ions
- Done by saprobionts

4) Nitrification - Ammonium ions to Nitrites to Nitrates
- Done by nitrifying bacteria

Question 6

The Phosphorus Cycle - in 7 steps

Answer 6

1. Phosphate ions in rocks are released into the soil by weathering.
2. Phosphate ions are taken into the plants through the roots. Mycorrhizae greatly increase the rate at which phosphorus can be assimilated.
3. Phosphate ions are transferred through the food chain as animals eat the plants and are in turn eaten by other animals.
4. Phosphate ions are lost from the animals in waste products.
5. When plants and animals die, saprobionts are involved in breaking down the organic compounds, releasing phosphate ions into the soil for assimilation by plants. These microorganisms also release the phosphate ions from urine and faeces.
6. Weathering of rocks also releases phosphate ions into seas, lakes and rivers. This is taken up by aquatic producers, such as algae, and passed along the food chain to birds.
7. The waste produced by sea birds is known as guano and contains a high proportion of phosphate ions. Guano returns a significant amount of phosphate ions to soils (particularly in coastal areas). It is often used as a natural fertiliser.

Question 7

IAA in phototropism in 3 steps

Answer 7

1) IAA is produced in the tips of growing shoots and roots and it diffuses to the shaded part of the shoot so theres uneven growth

2) This causes the shoot to bend towards the sun as there is cell elongation as cell walls break and stretch

3) IAA diffuses to the shaded side of the roots and this inhibits growth so the root bends away from the light

Question 8

Pacinian corpuscles - in 3 steps

Answer 8

1) Lamellae are deformed and press on the sensory nerve ending
2) This causes stretch mediated sodium ion channels to open
3) Influx of sodium ions causes a generator potential and if it reaches the threshold, an action potential will occur

Question 9

How photoreceptors work - 3 steps

Answer 9

1) Light enters the eye, hits the photoreceptors and is absorbed by light-sensitive optical pigments.

2) Light bleaches the pigments, causing a chemical change and altering the membrane permeability to sodium ions.

3) A generator potential is created and if it reaches the threshold, a nerve impulse is sent along a bipolar neurone which takes impulse to the brain

Question 10

Control of heart beat - in 3 steps

Answer 10

1) SAN acts as a pacemaker and it sends a wave of electrical activity causing both atria to contract

2) AVN gets the impulse and delays it by letting atria fully contract and empty before it allows the ventricles to contract

3) Wave of electrical activity goes through the Bundle of His and down the Purkyne Fibres which make sure the ventricles contract from the base upwards

Question 11

Control of heart rate in response to different stimuli - High Blood Pressure - in 3 steps

Answer 11

1) Baroreceptors detect high blood pressure which sends an impulse via sensory neurone to the medulla

2) The medulla send a impulse via the parasymapathetic neurone which also releases acetycholine

3) Acetycholine binds to the receptors on the SAN which reduces the heart rate and reduces blood pressure to normal

Question 12

Control of heart rate in response to different stimuli - Low Blood O2 - in 3 steps

Answer 12

1) Chemoreceptors detect low blood O2 which sends an impulse via sensory neurone to the medulla

2) The medulla send a impulse via the symapathetic neurone which also releases noradrenaline

3) Noradrenaline binds to the receptors on the SAN which increases the heart rate and increases blood oxygen back to normal

Question 13

How resting potential is maintained - in 3 steps

Answer 13

1) Sodium-Potassium pump, pumps out 3 Sodium ions for every 2 Potassium ions that come in

2) Potassium also diffuses out of the membrane via potassium ion channel via facilitated diffusion

3) This means that inside of the membrane is negatively charged compared to the outside of the membrane

Question 14

Action Potenial - in 5 steps

Answer 14

1) Stimulus - Sodium ion channels open making the membrane more permeable to sodium ions so sodium ions flow into the membrane making the inside of the membrane less negative

2) Depolarisation - If the voltage reaches the threshold an action potential will fire and more sodium ion channels will open and more sodium ions will flood in making the inside of the membrane positive relative to the outside

3) Repolarisation - The sodium ion channels close and the potassium ion channels open and the potassium ions flood out of the membrane making the inside of the membrane negative relative to the inside

4) Hyperpolarisation - The potassium ion channels are still closing so potassium ions are still flooding out and there is a slight overshoot making the voltage lower than at rest

5) Resting Potential - The potassium ion channels are close and the membrane goes back to 3 Na pumped out via active transport and 2 K in but facilitated diffusion of potassium ions out

Question 15

Cholinergic synapses - in 5 steps

Answer 15

1) Action potential stimulates voltage-gated calcium ion channels to open and allows calcium to flood into the synaptic knob

2) The influx of calcium ions causes vesicles in the synaptic knob to fuse to the presynaptic membrane

3) The vesicles release neurotransmitter acetycholine that diffuse across the synaptic cleft and bind to cholinergic receptors on the post synaptic membrane.

4) This causes sodium ion channels to open and sodium ion flood in depolarising the membrane and if threshold is met an action potential will fire

5) Acetycholine is removed from the receptors so the response doesn’t keep happening or AchE breaks it down so the products can go back to presynaptic neurone

Question 16

The process of muscle contraction - in 7 steps

Answer 16

1) Action Potential from motor neurone stimulates a muscle cell and causes depolarisation of the sarcolemma which spreads to the sarcoplasmic reticulum

2) The sarcoplasmic reticulum releases calcium ions into the sarcoplasm and the influx of calcium ions causes muscle contraction

3) Calcium ions bind to troponin changing its shape and causing it move tropomyosin and expose the actin-myosin binding site

4) Myosin head binds to actin filament to build an actin-myosin cross bridge and Calcium activates ATP hydrolase so ATP is hydrolysed to provide energy to make the myosin head to bend and the actin filament to follow the head.

5) Another ATP provides energy for the actin-myosin cross bridge to break and allowing another one to form at the new actin myosin binding site

6) When muscle has stop being stimulated, calcium ions leave their binding sites and moved back into sarcoplasmic reticulum via active transport via ATP.

7) Tropomyosin moves back and sarcomere lengthens

Question 17

Insulin - what it does - in 5 steps

Answer 17

1) Insulin is secreted by beta cells in the islets of Langerhans by the pancreas

2) Insulin binds to specific receptors on liver and muscle cells and increases their permeability to glucose so they take up more more glucose through more channel proteins

3) Insulin activates enzymes that covert glucose into glycogen - glycogenesis

4) Insulin increases respiration of glucose in muscle cells especially

5) Insulin reduces blood glucose concentration when its too high

Question 18

Glucagon - what it does - in 5 steps

Answer 18

1) Glucagon is secreted by alpha cells in the islets of Langerhans by the pancreas

2) Glucagon activates enzymes that convert glycogen into glucose - glycogenolysis

3) Glucagon secretes enzymes that convert amino acids and glycerol into glucose - gluconeogenesis

4) Glucagon decreases respiration of glucose in muscle cells especially

5) Glucagon increases blood glucose concentration when its too low

Question 19

Rise in blood glucose concentration - in 3 steps

Answer 19

1) The rise in blood glucose concentration is detected by the pancreas and its beta cells secrete insulin and alpha cells stop secreting glucagon

2) Insulin binds to specific receptors on the liver and muscle cells and they activate glycogenesis

3) Blood glucose concentration returns to normal

Question 20

Second Messengers - Adrenaline and glucagon - 2 steps

Answer 20

1) Adrenaline and glucagon bind to their receptors and activate an enzyme called adenylate cyclase which converts ATP into cAMP

2) cAMP activates an enzyme called protein kinase A which activates a cascade that breaks down glycogen into glucose (glycogenolysis)

Question 21

Ultrafiltration - in 4 steps

Answer 21

1) Blood from the renal artery passes through smaller arterioles in the cortex

2) The blood flows into the afferent arteriole and then through the efferent arteriole which forces out molecules as it puts it under pressure from the smaller diameter and forces small molecules into the Bowman’s Capsule

3) The filtered blood passes through the bowmans capsule where larger molecules like proteins and blood cells can’t pass through as it goes through the bowman’s capsule epithelium, the basement membrane and the capillary endothelium

4) This is now called the glomeular filtrate and useful substances are reabsorbed along the way and it passes through the collecting duct

Question 22

Selective Reabsorption - in 3 steps

Answer 22

1) The proximal convoluted tubule has microvilli to increase surface are so useful solutes like glucose can be reabsorbed via active transport and facilitated diffusion

2) Water moves from filtrate to blood via osmosis as blood has a lower water potential

3) Water is reabsorbed from the convoluted tubules, the collecting duct and the Loop of Henle and the rest is urine

Question 23

Loop of Henle in 5 steps

Answer 23

1) Sodium ions actively transported into the medulla reducing its water potential at the top of the ascending limb

2) Water move into the medulla from the filtrate in the descending limb via osmosis

3) Sodium ions diffuse out into the medulla at the bottom of the ascending limb reducing water potential of medulla

4) Water from the distal convoluted tubule and collecting duct move into the medulla via osmosis.

5) All water in the medulla is reabsorbed in the blood via capillary networks

Question 24

Osmoregulation in 5 steps

Answer 24

1) Low WP is detected by osmoreceptors in the hypothalumus which causes them to lose water via osmosis

2) This send a signal that is picked up by other cells in the hypothalumus which send a signal to the posterior pituritary gland which secretes ADH

3) ADH bind to receptors on the cell membranes of the Distal Convoluted Tubule and the collecting duct and it inserts aquaporins which increase permeability

4) More water is reaborbed from the tubule to the medulla to the blood

5) Small amount of highly concentrated urine is produced so less water is lost

Question 25

Allopatric Speciation - in 4 steps

Answer 25

1) Reproductive isolation due to geographical seperation via physical barrier 2) The populations will experience different selection pressures and so different changes in allele frequencies could occur 3) Different alleles will be more advantageous in the different populations, so natural selection occurs 4) They won’t be able to breed with one another to produce fertile offspring and they will be seperate species

Question 26

Sympatric Speciation - in 3 steps

Answer 26

1) Random mutations within a population prevent individuals that carry the mutation from breeding with other members of the population that don’t carry the mutation. 2) Reproductive isolation without geographical isolation 3) Results in not being able to breed together to produce fertile offspring so they are seperate species

Question 27

Random sampling - in 5 steps

Answer 27

1) Choose an area to sample-a small area within the area being investigated. 2) Samples should be random to avoid bias. You can use a random number generator to ensure your samples are random (see below). 3) Use an appropriate technique to take a sample of the population 4) Repeat the process, taking as many samples as possible. This will reduce the likelihood that your results are down to chance 5) The number of individuals for the whole area can then be estimated by taking the mean of the data collected in each sample and multiplying it by the size of the whole area. The percentage cover for the whole area can be estimated by taking the mean of all the samples.

Question 28

Succession - in 5 steps

Answer 28

1. The pioneer species are able to survive because they can survive harsh environments 2. Pioneer species change the abiotic conditions as they die and are decomposed helping to form a basic soil. This makes the environment less hostile 3. The new organisms then die and are decomposed, adding more organic material, making the soil deeper and richer in minerals such as nitrates 4. Shrubs, ferns and small trees begin to grow, out-competing the grasses and smaller plants to become the dominant species. Diversity increases. 5. Finally, the soil is deep and rich enough in nutrients to support large trees. These become the dominant species, and the climax community is formed.

Question 29

Eutrophication - in 5 + 1 steps

Answer 29

1) Algae blooms block light 2) No photosynthesis, so plants die. 3) Saprobiotic decomposition occurs. 4) Microorganisms respire aerobically (use up oxygen) 5) Less oxygen for fish to respire, therefore they die 6) Increase in anaerobic microorganisms; release hydrogen sulphide/ nitrates/ toxic waste

Question 30

The digestion of carbohydrates in 3 steps

Answer 30

1) Amylase works by catalysing hydrolysis reactions that break the glycosidic bonds in starch to produce maltose (a disaccharide). 2) Maltase (membrane-bound disaccharidase) breaks down maltose (disaccharide) into 2 alpha glucose (monosaccharide) by hydrolysing glycosidic bonds. 3) The alpha glucoses (monosaccharides) can be transported across the epithelial cell membranes in the ileum via specific cotransporter proteins alongside sodium ions

Question 31

The digestion of lipids in 3 steps

Answer 31

1) Bile salts are produced by the liver and emulsify lipids meaning they cause the lipids to form small droplets to increase the surface area for lipases to work on. 2) Lipase enzymes catalyse the breakdown of lipids into monoglycerides and fatty acids involving the hydrolysis of the ester bonds. 3) The monoglycerides and fatty acids stick with the bile salts to form micelles which help the products to be absorbed

Question 32

The digestion of proteins in 3 steps

Answer 32

1) Endopeptidases hydrolyse peptide bonds within a polypeptide 2) Exopeptidases hydrolyse peptide bonds at the ends of polypeptides removing single amino acids 3) Dipeptidases separate the two amino acids that make up a dipeptide by hydrolysing the peptide bond between them

Question 33

Water movement up a plant - Cohesion and tension (4 steps)

Answer 33

1. Water evaporates from the leaves at the 'top' of the xylem. This is a process called transpiration. 2. This creates tension (suction), which pulls more water into the leaf. 3. Water molecules are cohesive (they stick together) so when some are pulled into the leaf others follow. This means the whole column of water in the xylem, from the leaves down to the roots, moves upwards. 4. Water then enters the stem through the roots.

Question 34

The mass flow hypothesis (3 steps with examples throughout)

Answer 34

1. Source - Active transport is used to actively load the solutes (e.g. sucrose from photosynthesis) from companion cells into the sieve tubes of the phloem at the source (e.g. the leaves). - This lowers the water potential inside the sieve tubes, so water enters the tubes by osmosis from the xylem and companion cells. - This creates a high pressure inside the sieve tubes at the source end of the phloem. 2. Sink - At the sink end, solutes are removed from the phloem to be used up. - This increases the water potential inside the sieve tubes, so water also leaves the tubes by osmosis. - This lowers the pressure inside the sieve tubes. 3. Flow - The result is a pressure gradient from the source end to the sink end. - This gradient pushes solutes along the sieve tubes towards the sink. When they reach the sink the solutes will be used (e.g. in respiration) or stored (e.g. as starch). - The higher the concentration of sucrose at the source, the higher the rate of translocation.

Question 35

The main stages of the immune response - 1. Phagocytosis What is phagocyte, an example and where is it found? How does it work?

Answer 35

1. Phagocytosis A phagocyte (e.g. a macrophage) is a type of white blood cell that carries out phagocytosis (engulfment of pathogens). They're found in the blood and in tissues and are the first cells to respond to an immune system trigger inside the body - A phagocyte recognises the foreign antigens on a pathogen. - The cytoplasm of the phagocyte moves round the pathogen, engulfing it. - The pathogen is now contained in a phagocytic vacuole in the cytoplasm of the phagocyte. - A lysosome fuses with the phagocytic vacuole. The lysozymes break down the pathogen. - The phagocyte then presents the pathogen's antigens - it sticks the antigens on its surface to activate other immune system cells. The phagocyte is acting as an antigen-presenting cell.

Question 36

The main stages of the immune response - 2. T-Cells What is a T Cell? What does it do? What are the different types of T Cells? What do T Cells also do?

Answer 36

2. T-Cells A T-cell (also called a T-lymphocyte) is another type of white blood cell. - It has receptor proteins on its surface that bind to complementary antigens presented to it by phagocytes. This activates the T-cell. Different types of T-cells respond in different ways. For example: - Helper T-cells (TH cells) release chemical signals that activate and stimulate phagocytes and cytotoxic T-cells (T cells), which kill abnormal and foreign cells. - T cells also activate B-cells, which secrete antibodies

Question 37

The main stages of the immune response - 3. B-Cells What is a B-cell? What is different about every antibody? Clonal Selection?

Answer 37

B-cells (also called B-lymphocytes) are also a type of white blood cell. They're covered with antibodies-proteins that bind to antigens to form an antigen-antibody complex. Each B-cell has a different shaped antibody on its membrane, so different ones bind to different shaped antigens When the antibody on the surface of a B-cell meets a complementary shaped antigen, it binds to it. This, together with substances released from helper T-cells, activates the B-cell. This process is called clonal selection. The activated B-cell divides into plasma cells.

Question 38

The main stages of the immune response - 4. Antibody production What are plasma cells to the B-cell? What do they do? What are these called and what do they do? What does the antibody have and do? What do phagocytes then do? What are antibodies and what do they have? Figure 5: Antigen-antibody complex and antibody structure.

Answer 38

4. Antibody production Plasma cells are identical to the B-cell (they're clones). They secrete loads of antibodies specific to the antigen. These are called monoclonal antibodies. They bind to the antigens on the surface of the pathogen to form lots of antigen-antibody complexes . An antibody has two binding sites, so can bind to two pathogens at the same time. This means that pathogens become clumped together - this is called agglutination. Phagocytes then bind to the antibodies and phagocytose many pathogens at once. This process leads to the destruction of pathogens carrying this antigen in the body. Antibodies are proteins—they're made up of chains of amino acids. The specificity of an antibody depends on its variable regions, which form the antigen binding sites. Each antibody has a variable region with a unique tertiary structure (due to different amino acid sequences) that's complementary to one specific antigen. All antibodies have the same constant regions.