| The kidney function | - Osmoregulation
- Excretion - to remove nitrogenous waste e.g. urea |
| Osmoregulation stages | - The formation of glomerular filtrate by ultrafiltration
- Reabsorption of glucose and water by the proximal convoluted tubule
- Maintenance of a gradient of sodium ions in the medulla by the loop of Henle
- Reabsorption of water by the distal convoluted tubule and collecting ducts |
| What is osmoregulation? | - The maintenance of water potential and blood plasma and tissue fluid |
| What causes a build-up of hydrostatic pressure within the glomerolus? What is the result of this? | - The afferent arteriole is larger in diameter than the efferent arteriole
- Water, mineral ions and glucose are squeezed out of the capillaries to form glomerular filtrate
- Blood cells and proteins are too large to pass into the Bowman's capsule from the afferent arteriole |
| What five factors causes resistance to the formation of glomerulus filtrate? | 1) Capillary endothelial cells
2) Epithelial cells of the Bowman's capsule
3) Connective tissue and epithelial cells of the blood capillary
4) Hydrostatic pressure of the fluid in the Bowman's capsule space
5) Low water potential of the blood in the glomerulus |
| What are the two modifications that reduce the resistance of the formation of the glomerulus filtrate? | 1) Bowman's capsule made up of cells called podocytes, space between them allows fitrate to pass through
2) The endothelium of the glomerular capillaries have spaces between cells |
| Why is filtrate formed, even against all the resistance? | - The hydrostatic pressure is great enough to overcome the resistance
- Filtrate passes out from the blood into the Bowman's capsule |
| How does ultrafiltration work? | - Afferent arteriole is wider than the efferent arteriole creating a high hydrostatic pressure
- Small molecules h20, glucose, ions, urea is filtered through the capillaries, basement membrane, through slit pores into bowmans capsule (filtrate)
- Large molecules (plasma proteins) remain in the plasma |
| What three structures do the epithelial cells in the proximal convoluted tubule have and why does it make them well adapted for reabsorbtion? | 1) Microvilli- provide large surface area of membrane - more space for protein carriers - facilitated diffusion
2) High density of mitochondria to provide ATP for active transport
3) There is only 1 layer of cells between the PCT + the capillary - short diffusion distance |
| Describe the four steps of the process of reabsorption into the proximal convoluted tubule (1) | - In the glomerular filtrate, glucose + sodium ions will move down conc grad. into the cell via cotransport across a carrier protein
- This mov. of ions will lower water potential in cell |
| Describe the four steps of the process of reabsorption into the proximal convoluted tubule (2) | - Causing h20 to move down the h2o potential gradient into cell via osmosis
- In the back of the cell, sodium ions will be AT to the inter cell space, always establishing a conc grad of sodium ions |
| Describe the four steps of the process of reabsorption into the proximal convoluted tubule (3) | - The sodium ions will then diffuse into the capillary by simple diffusion
- Glucose moves by facilitated diffusion into intermembrane space, then moves into cap by simple diffusion |
| What is the loop of Henle responsible for? | - Provides osmotic gradient in medualla allowing reabsorption of water from the collecting duct
- Acts as a counter-current multiplier |
| What is it important that water is reabsorbed from the collecting duct? | - To concentrate the urine and lower its water potential to below that of the blood |
| What are the two regions of the loop of Henle? | - The descending limb
- The ascending limb |
| Process at the loop of Henle (1) | - The descending limb is permeable to water
- But its impermeable to na ions
- So water can leave it but ions cant |
| Process at the loop of Henle (2) | - h20 leaves filtrate, its water potential is lowered as loop descends into the medulla
- As the loop ascends na ions leave the ascending limb initially by facilitated diffusion |
| Process at the loop of Henle (3) | - The upper part of the ascending limb na ions is removed by AT
- The removal of na ions leads to increase in h2o potential as filtrate passes up ascending limb |
| Process at the loop of Henle (4) | - Filtrate passes down collecting duct
- h2o potential around duct decreases ass it descends
- So water leaves collecting duct by osmosis
- Then h2o enters the capillaries by osmosis |
| How does water pass out of the collecting duct? | - By osmosis
- Through channel proteins that are specific to water, called aquaporins |
| What does ADH do? | - It alters the number of channel proteins and thus, controls water loss |
| What are the two main roles of the distal tubule? | - To make final adjustments to the water and salts that are reabsorbed
- Control the pH of the blood by selecting which ions to reabsorb |
| Give an example of how counter-current is used within the loop of Henle and why this is useful | - The filtrate in the collecting duct with a low water potential meets interstitial fluid with a lower water potential
- Although there is a small water potential gradient, it exists for the whole length of the collecting duct
- This allows for a steady flow of water into the interstitial fluid and thus, blood |
| What would happen if the two flows were in the same direction, parallel? | - Less of the water would enter the blood |
