Paper 2 Mistakes Flashcards
(43 cards)
Explain how nitrate might cause the death of fish in fresh water
Growth of algae / surface plants / algal bloom blocks light;
Reduced / no photosynthesis so (submerged) plants die;
Saprobiotic (microorganisms / bacteria);
Aerobically respire / use oxygen in respiration;
Less oxygen for fish to respire / aerobic organisms die;
Role of microorganisms in producing nitrates from the remains of dead organisms
- Saprobiotic (microorganisms / bacteria) break down remains / dead material / protein / DNA into ammonia / ammonium;
- Ammonia / ammonium ions into nitrite and then into nitrate;
- (By) Nitrifying bacteria / nitrification;
Ultrafiltration
- High hydrostatic pressure due to afferent arteriole having larger diameter than efferent
- Capillary wall made of endothelium, has fenestrations
- Allows soluble molecules (urea, glucose, amino acids, ions) and water to be forced out
- Basement membrane filters out small molecules
- Plasma proteins and RBC too large = remain in blood
- Podocyte allowsfiltrate to move into Bowman’s capsule
What molecules should be selectively reabsorbed
100% glucose and amino acids
80% water
Some ions
Part of kidney that PCT and DCT are in
Medulla
What happens to water potential the further down the medulla you go?
Water potential becomes more negative
Due to more solutes
Describe the process of selective reabsorption in the proximal convoluted tubule
-blood in capillaries has low water & solute levels due to glomerular filtrate
-Na⁺/K⁺ pumps in basal membranes actively transport Na⁺ out of epithelial cells into capillaries
-lowers Na⁺ conc inside cells
-Na⁺ diffuses from filtrate into epithelial cells via co-transporter proteins
-brings glucose/amino acids with them
-glucose/amino acids then diffuse into blood through transport proteins in basal membrane
-lowers water potential of blood
-causes water to move into blood by osmosis
-higher conc of urea in filtrate than blood
-some urea is also reabsorbed by diffusion down its concentration gradient
📝 Markscheme tip: Must mention:
Na⁺/K⁺ pump
Co-transport with glucose/amino acids
Osmosis due to water potential gradient
Urea diffuses due to concentration gradient
How are cells of the proximal convoluted tubule adapted for selective reabsorption?
Microvilli increase surface area for reabsorption
Many mitochondria = lots of ATP for active transport
Co-transporter proteins allow Na⁺ + glucose/amino acids in
Basal membrane transport proteins allow glucose/amino acids to diffuse into blood
Tightly packed epithelial cells = shorter diffusion distance
Close contact with capillaries = short diffusion distance for solutes and water
What happens to water, salts, and urea in the nephron after the PCT?
In Loop of Henle:
Salts are reabsorbed via loop of Henle and collecting duct into blood by diffusion
Creates a water potential gradient
Water follows salts into blood by osmosis
In collecting duct:
Water reabsorption is regulated based on body’s needs
More water reabsorbed when dehydrated (ADH presence)
Urea:
Concentration in filtrate ↑ after water reabsorption
Some urea diffuses back into blood due to its higher concentration in filtrate
-after the necessary reabsorption of amino acids, water, glucose and inorganic ions is complete, the filtrate eventually leaves the nephron
-now referred to as urine
-flows out of the kidneys, along the ureters and into the bladder, where it is temporarily stored
What happens in dim light?
-circular muscles relax
-radial muscles contract
-allows pupil to dilate
-maximises light that can enter eye
What happens in bright light?
-pupil constricts to let minimal light in
-prevents any damage occurring
-radial muscle relax and circular muscles contract
Where are nephrons found?
Medulla of kidney
Nephron structure
-afferent arteriole leads into nephron
-branches into many small capillaries (glomerulus)
-glomerulus inside Bowman’s capsule
-leads to proximal convuluted tubules
-leads to loop of Henle
-leads to distal convoluted tubule
When and where is glucose reabsorbed?
Selective reabsorption in the PCT
Estruch ultrafiltration
-blood enters through afferent arteriole
-splits into lots of smaller capillaries, making up the glomerulus
-causes a high hydrostatic pressure in blood
-blood in the glomerular capillaries is separated from the lumen of the Bowman’s capsule by 3 layers of cells:
-first cell layer = capillary endothelium = each cell is perforated
-middle layer = basement membrane = made up of a network of collagen and glycoproteins
-second cell layer = epithelium of Bowman’s capsule (cells have many tiny finger-like projections with gaps in between them called podocytes)
-blood passes through the glomerular capillaries
-holes in the capillary endothelial cells and the gaps between the podocytes allows substances dissolved in the blood plasma to pass into the Bowman’s capsule
-water and small molecules, e.g. glucose/mineral ions are forced out of the capillaries
-form the glomerulus filtrate
-large proteins and blood cells are too big
-cannot fit through gaps in the capillary endothelium
-remain in the blood, which leaves via the efferent arteriole
Estruch selective reabsorption
Selective reabsorption:
-Na+ ions are actively transported out of PCT epithelial cells into blood in capillaries
-lowers concentration of Na+ ions in the PCT epithelial cell
-Na+ ions diffuse down the gradient from the lumen of the PCT into the cells lining the PCT (via co-transport, carrying glucose with it)
-glucose can diffuse from the PCT epithelial cell into the blood stream
Tip: similar to co-transport for digestion/absorption in Module 3
Estruch loop of Henle
-Na+ ions are actively transported out of the ascending limb of the loop of Henle
-uses energy from ATP produced in mitochondria in the walls of the cells
-Na+ ions accumulate outside of the nephron in the medulla
-lowers the water potential in interstitial space
-ascending limb is impermeable to water
-so water diffuses out of the descending limb by osmosis into the interstitial space and then the blood capillaries
-hence water is reabsorbed into the blood
-at the base of the ascending limb some Na+ ions diffuse out
-as there is now a very dilute solution due to all the water that has moved out
3 layers that separate blood in the glomerular capillaries from the lumen of the Bowman’s capsule
-first cell layer = capillary endothelium = each cell is perforated
-middle layer = basement membrane = made up of a network of collagen and glycoproteins
-second cell layer = epithelium of Bowman’s capsule (cells have many tiny finger-like projections with gaps in between them called podocytes)
Line transects method
Place the tape measure at a right angle to the shore line
Place the quadrat at every 5 metres / every position
Collect the data (percentage cover or local frequency or density)
Repeat by placing another 30 transects along the beach at right angles to the shore line
Always be specific about where you place it
Methods to estimate the abundance of a species
- Local frequency = % of squares in the quadrat with the species present
- Density: the number of one species in a given area, then multiply to work out number in whole field
- Percentage Cover: proportion of the ground occupied by the species
Pros and cons of each method
Local frequency:
-quick method, especially if too many to count
-inaccurate, doesn’t consider overlapping plants or size of the plant
Density:
-more accurate if you can distinguish individual plants
-can be used to estimate species richness
-but more time-consuming
Percentage cover:
-quicker, useful if hard to identify individual organism or too many to count
-subjective estimate = inaccurate
-doesn’t consider overlapping plants or size of plants
Mark Release Re-capture method
- Initial sample of population is caught
- Mark these individuals and release back into the wild and note down the number
- Allow individuals to randomly disperse, then take second sample (e.g. 24 hours later)
- Record the number caputred in the second sample, and how many ofnthem are already marked
- Estimate size
Number of organisms in first sample multipled by number of individuals in second sample
Then divide by number of marked organisms
Assumptions for mark-release-recapture
Population is constant (no birth/death/migration)
Animals always redistribute evenly (they may all huddle near food/water source/shelter)
Evidence from pedigree diagram that condition is caused by recessive allele
-both parents are white, so they don’t have the condition
-child is black, so has the condition
-hence both parents must be carriers of the recessive allele
-but due to the presence of dominant allele, they don’t have the condition
