Application Flashcards
scientists investigated the movement of iron ions (Fe3+) from the soil to old and young leaves of heat-treated barley plants and to leaves of plants that were not heat treated.
heat treatment was applied half way up to the leaves. the scientists determined the concentration of Fe3+ in the top and lower halves of the leaves of each plant.
Figure showed old leaf being similar in both heat treatment and untreated showed young leaf having lower ratio in treated than untreated.
what can you conclude about the movement of Fe3+ in barley plants?
- heat treatment has greater effect on young leaves than old
- heat treatment damages phloem
- Fe3+ moves up leaf
- Fe3+ transported in xylem in older leaf
- in young leaf, some in xylem, as some still reaches top part of leaf
- all ratios show there is less Fe3+ in old leaves than yong
- no statistical test to show significant difference
genus
spores spread
- measure diameter of large number of spores
- divide measured values by 700 to get actual diameter
- using volume of sphere
explain the shape of curves between 50 and 60 degrees
- denaturation at both
- at 60 is faster - more KE
- active changes shape - no ESC
- hydrogen bonds broken
- rapid digestion of proteins
- can absorb more amino acids for growth, reprod etc
- hydrolyse peptide bonds to rel amino acids
- amino acids can cross cell surface membrane
kingdom - mammalia
phylum
class
order
genus - cavia
species - porcellus
what limits the rate of this reaction between points A and B
give evidence from graph
- substrate concentration
- as substrate concentration inc, rate inc
suggest a reason for the shape of the curve between C and D
all active sites filled up
- similar shape
- less ESC formed
explain why doctors stopped measuring the amounts of 14C
explain how 14C in CO2 came from 14C in glucose in the blood?
- little or no difference at 8 hours
- between ALL groups
CO2 in the breath contained in the radioactive form of carbon, 14C
explain how 14C in carbon dioxide came from 14C in glucose in the blood
- respiration occurs
- by cells/tissues
doctors concluded that measuring the amount of 14C in the carbon dioxide in the breath after 3 hours was a better way of diagnosing lactose deficiency than the lactose intolerance test
do you agree
- clear difference
- between lactose deficient and IBS group
- as SDs dont overlap
- greater rate of oxygen consumption - greater rate of resp - greater rate of uptake
- oxygen needed for resp
- respiration produces ATP
- K+ uptaken by active transport which needs ATP
explain why 2cm3 of water was added to tubes A and C
- give same vol of B
- to keep conc of gelatine constant
explain results of A and B
- enzyme in pineapple digested gelatine
- no gelatine - no jelly
- enzyme denatured due to hydrogen bonds, ionic and disulfide bonds broken
- by HCl - due to change in pH
what is the purpose of tube c
- for comparison / control
- show enzyme in pineapple that digested gelatine boiling denatured
describe how you would expect the number of lysosomes in a pupa to change with age of pupa
- fall, increase, fall
- lysosomes associated with tissue breakdown
suggest an explanation for the change in RNA conc in the first 40% of time spent in pupa
- tissue cells being broken down
- RNA hydrolysed
- by enzymes from lysosomes
- no new RNA made
suggest an explanation for the change in RNA conc between 60 and 80 % of time spent as a pupa
- RNA associated with making protein
- new adult tissue forming
cytoplasm of red blood cells filled with haemoglobin
- membrane has phospholipid bilayer
- stain binds to phospholipid
- on inside AND outside of membrane
- high to low using carrier / channel proteins
- specific to substrate
- down conc gradient
explain difference between curve B and C
- with comp inhib - enzyme sub collisions still inc
- binding of comp - no change in shape of active
- binding on non-comp - changing of active site
- at high sub conc - enzyme active site available
- at high sub conc with non comp - not available
- repeat soil sorting for diff times and record number of species collected
- find optimum time beyond which no further sorting leads to inc in animal species found
- little digestion of starch by salivary amylase
- amylase also secreted by pancreas
- most starch digestion occurs in small intestine
- hydrogen bonds between pairs holds pairs together
- MANY hydrogen bonds provide strength
ribosomes assemble polypeptides using mRNA code
OR
DNA has 2 strands each with different base sequence
- stops translation
- allows detachment of polypeptide from ribosome
why grew in nutrient poor sand
no boron provided
why grew in dark
no sucrose produced in leaves by photosynth
- evaporate all water
- no burning
do results support scientists hypothesis
YES
- uptake of sucrose greater
- transport to other parts of plant
NO
- no evidence that boron reacts with sucrose
- take thin section of plant tissue
- place against photographic film in dark for several hours
- antigen on surface of Nm binds to surface of protein on Bcell
- B cell divides by mitosis
- division stimulated by T cells
- plasma cells release antibodies
- B cells become memory cells
- memory cells produce antibodies faster
- mutation
- results in resistancy allele in Nm to ONE antibiotic
- cell survives, reproduces and passes allele for resistancy onto offpsring
- process repeated with diff genes - resistant to 2 antibiotics
