Progression kRQ Flashcards
What is digestion?
The hydrolysis of large biological molecules into smaller molecules that can be absorbed across the cell membrane of epithelial cells in the lumen (small intestine)
Describe and explain how carbohydrate digestion occurs in mammals?
Starch digestion begins in the mouth when salivary amylase from salvia mixes with food in the mouth. Salivary amylase hydrolyses glycosidic bonds turning some starch into maltose.Food is swallowed and enters the stomach where salivary amylase is denatured due to low pH. Pancreatic amylase is released into the small intestine along with alkaline bile slats from the liver to neutralise the acid. Pancreatic amylase hydrolyses the remaining starch into maltose. In the cell membranes of the epithelial cells lining the lime are membrane bound disaccharides, maltose, lactase,sucrase. These hydrolyse disaccharides into monosaccharides for absorption
What is the advantage of membrane bound enzymes?
They do not need to be constantly replenished as they are not removed with waste from the small intestine because they are bound to the cell surface membrane of the epithelial cell membrane,so have a very short diffusion distance into the cell
What are the membrane bound enzymes and what are their substates and products
Sucrase-hydrolyses sucrose into fructose and glucose
Maltose-hydrolyses maltose into glucose
Lactase-hydrolyses lactose into galactose and glucose
Dipeptidase-hydrolyses dipeptides into amino acids
Describe how proteins are digested into the human gut endorses produced by the stomach.
Endopeptidases produced by the stomach. Endopeptidases are also made by the pancreas and are released into the small intestine. Endopeptidases hydrolyse internal peptide bonds which breaks polypeptides into smaller polypeptide chains.
Exopeptidases are made by the pancreas and released into the small intestine. They hydrolyse peptide bonds at the ends of the polypeptide chain releasing amino acids or dipeptides.
Dipeptidases hydrolyse dipeptides into amino acids.
- What are the advantages of using endopeptidases and exopeptidases
Endopeptidases hydrolyse internal peptide bonds therefore there are more ends or increase in surface area for exopeptidases which hydrolyse bonds at ends of chains.
- Describe how glucose/amino acids are absorbed by cotransport.
Sodium ions actively transported from ileum/epithelial cell to blood; This forms a concentration gradient for sodium to enter cells from gut lumen and with it, glucose/amino acids enters by facilitated diffusion with sodium ions against the glucose/amino acid concentration gradient; Glucose/amino acid diffuses into the blood, down the concentration gradient, through another protein channel.
- Explain why cotransport of glucose and amino acids is called indirect active transport.
Energy is used to actively transport the sodium ions out of the epithelial cell into the blood, glucose/amino acids are transported passively at the other end of the cell.
- Describe and explain how lipids are digested in the human gut.
Bile salts emulsify lipids from large droplets to small droplets, small lipid droplets have a large surface area for lipase to hydrolyse ester bonds forming monoglycerides and fatty acids. The monoglycerides and fatty acids remain bound to the bile salts in small droplets called micelles.
- Explain why the pH decreases when lipids are hydrolysed.
Fatty acids are produced, which are acidic and so lower the pH.
- Describe and explain how lipids are absorbed.
Micelles interact with the phospholipid bilayer of the epithelial cells and the fatty acids/monoglycerides enter by simple diffusion as they are lipid soluble. Once in the cell triglycerides are reformed by the smooth endoplasmic reticulum, then the golgi modifies the triglyceride but forming structures called chylomicrons (triglycerides, cholesterol, protein and phospholipids) the golgi packages chylomicrons into vesicles. Vesicles perform exocytosis releasing chylomicrons into lacteals (lymph system). The lymph system then carries the chylomicrons to the cardiovascular system.
- What are the benefits of small droplet and micelle formation?
Droplets increase surface areas for lipase action; So faster hydrolysis of lipids; Micelles make the fatty acids and monoglycerides more soluble in water and carry fatty acids and monoglycerides to the cell membrane of intestinal epithelial cell wo they can be absorbed.
- State two functions of bile salts
Emulsification of lipids into small droplets. Neutralisation of hydrochloric acid.
- Describe the cohesion-tension theory of water transport in the xylem.
Water evaporates from leaves, lowers water potential of leaf cells; Water pulled up xylem creating tension; Water molecules are cohesive due to hydrogen bonds between water molecules; forming a continuous water column. Adhesion of water molecules to walls of xylem also occurs; reducing the trunk diameter.
- Describe the mass flow hypothesis for the mechanism of translocation in plants.
At source sucrose is actively transported into the phloem;
By companion cells; this lowers water potential in phloem and water enters by osmosis; This produces high hydrostatic pressure; Causing the mass flow towards sink tissue; At sink sucrose is removed; and used in respiration or stored as starch, water moves out by osmosis so the hydrostatic pressure is lower. Mass flow is down a hydrostatic pressure gradient.
- Explain how sieve cells are adapted for mass transport.
They have few organelles and very little cytoplasm; So easier flow. End walls have perforations to allow movement of substances between cells.
- Explain how companion cells are adapted for mass transport.
Mitochondria release energy / ATP; For active transport.
Ribosomes produce proteins; for carrier proteins for active loading of sucrose into sieve tube elements.
- Explain how xylem is adapted for mass transport.
Xylem vessels have no end walls so form a series of unbroken tubes from root to leaf. Cells are dead so no cytoplasm or organelle which would restrict flow. Spirals of lignin withstand changes in pressure and prevent collapsing of the tube. Pits in the lignin allow lateral movement of water around blocked vessels.
- Describe and explain the precautions you should take when using a photometer, to obtain reliable results.
- Seal joints to ensure watertight; 2. Cut shoot under water to prevent air entering the xylem;
- Cut shoot at a slant; 4. Dry off leaves to increase the diffusion gradient; 5. Insert into apparatus under water to prevent air getting into the xylem; 6. Ensure no air bubbles are present;
- Shut tap; 8. Note where bubble is at start to find the distance it moves;
- A potometer measures the rate of water uptake rather than the rate of transpiration. Give two reasons why the potometer does not truly measure the rate of transpiration.
Water used for support of plant; Water used in photosynthesis; Water produced in respiration;
- Explain why transpiration rate is highest around midday.
There is more photosynthesis at midday as there is more sunshine, so rate of water uptake will be higher. It is usually warmer, so more evaporation will take place, increase transpiration rate.
- What does a ringing experiment show about translocation?
The phloem is involved in translocation, (bulge at top of ring) shows mass flow down the stem.
- Why can radioactively labelled CO2 be used to investigate translocation.
The CO2 is made into glucose/sucrose in the leaf via photosynthesis, then transported round the plant in translocation. It can be seen using an x-ray of the plant.
- Sketch and describe an oxygen dissociation curve.
Partial pressure of oxygen on the x-axis, percentage saturation of haemoglobin on the y-axis up to 100%. A S-shaped curve, reaching plateau just below 100%
- Use an oxygen dissociation curve to describe how haemoglobin loads and unloads oxygen in the body.
Haemoglobin Loading takes place at high pp.O2 (in the lungs); In lungs haemoglobin is almost fully saturated as it has a high affinity for oxygen; Haemoglobin Unloads or dissociates oxygen at low pp.O2 (at the respiring cells). The higher carbon dioxide concentration at cells further decreases the affinity of haemoglobin for oxygen (Bohr affect).
- Why is haemoglobin considered a quaternary structure?
Due to the presence of 4 polypeptide chains. It also has four prosthetic groups (Fe2+).