A level Biology Paper June 2021 Flashcards
(19 cards)
Describe the induced-fit model of enzyme action and how an enzyme acts as a
catalyst.
Substrate binds to active site
Active site changes shape so its complementary to substrate
enzyme substrate complexes formed
lowers activation energy
Suggest and explain a procedure the scientists could have used to stop each reaction
boil
enzymes denature
Explain the change in ATP concentration with increasing inorganic phosphate
concentration.
- (With) increasing Pi concentration, more
enzyme-substrate complexes are formed; - At or above 40 (mmol dm-3) all active sites
occupied
Explain the advantage for larger animals of having a specialised system that
facilitates oxygen uptake.
larger animals have a smaller surface area to volume ratio
so diffusion pathway is shorter so rate of diffusion is faster
Explain how the counter-current principle allows efficient oxygen uptake in the
fish gas exchange system.
- Blood and water flow in opposite directions;
- Diffusion/concentration gradient (maintained)
along (length of) lamella/filament;
Describe how one amino acid is added to a polypeptide that is being formed at a
ribosome during translation.
tRNA brings specific amino acid to ribosome
anticodon on tRNA complementary to codon on mRNA
amino acids join together via condensation reaction by ATP
amino acids join together to form a peptide bond using ATP
Crystallin is a structural protein found in the human eye. An inherited disease that
leads to blindness is caused by changes in properties of crystallin. The replacement
of the amino acid Arg with the amino acid Gly causes these changes.
Use information in Table 2 to suggest why this amino acid replacement changes the
properties of crystallin
Hydrogen bonds form instead of ionic
alters/changes the tertiary structure
Name the three phases of mitosis shown by C, D and E on Figure 7.
Describe the role of the spindle fibres and the behaviour of the chromosomes during
each of these phases
C = prophase and
D = metaphase and
E = anaphase;
- (In) prophase, chromosomes condense;
- (In) prophase OR metaphase, centromeres
attach to spindle fibres; - (In) metaphase, chromosomes/pairs of
chromatids at equator/centre of spindle/cell; - (In) anaphase, centromeres divide;
- (In) anaphase, chromatids (from each pair)
pulled to (opposite) poles/ends (of cell); - (In) prophase/metaphase/anaphase, spindle
fibres shorten;
Use your knowledge of phagocytosis to describe how an ADC enters and kills the
tumour cell.
- Cell ingests/engulfs the antibody/ADC
- Lysosomes fuse with vesicle/phagosome
(containing ADC); - Lysozymes breakdown/digest the antibody/ADC
to release the drug;
Some of the antigens found on the surface of tumour cells are also found on the
surface of healthy human cells.
Use this information to explain why treatment with an ADC often causes side effects.
- ADC will bind to non-tumour/healthy cells;
- Cause death/damage of non-tumour/healthy
cells
Suggest one reason why there are no data for Group G and Group H after day 8
mice died
Suggest and explain two further investigations that should be done before this ADC
is tested on human breast cancer patients.
test on other mammals to check for side effects
test on healthy humans to check for side effects
Describe how a triglyceride molecule is formed
3 fatty acids and 1 glycerol
condensation reaction and removal of 3 water molecules
ester bonds formed
The scientists used a data logger to measure the length of the root rather than a ruler.
Suggest one reason why they used a data logger and explain why this was important
in this investigation.
to increase accuracy because differences are small
It is known that:
- during respiration saturated fatty acids yield more energy than unsaturated fatty
acids - saturated fatty acids have higher melting points than unsaturated fatty acids
- lipases in seeds act more rapidly on liquid substrates.
Use this information and Table 6 to show how each population is better adapted for its
natural environment when compared with the other population.
- .Population 1 grew longer roots in warm
temperatures and population 2 grew longer
roots in cool temperatures; - Standard deviations do not overlap so
difference (in mean) unlikely to be/not due to
chance; - Population 1 (is better adapted to warm
conditions because it) has more saturated fatty
acids so more energy available (and more
growth); - Population 2 (is better adapted to cool
conditions because it) has more
unsaturated/liquid fatty acids so more lipase
activity (and more growth);
Although these two populations are completely separate and show genetic variation,
they are both called Helianthus annuus.
Explain why they are both given this name.
same genus
Describe the structure of DNA.
- Polymer of nucleotides;
- Each nucleotide formed from deoxyribose, a
phosphate (group) and an organic/nitrogenous base; - Phosphodiester bonds (between nucleotides);
- Double helix/2 strands held by hydrogen bonds;
- (Hydrogen bonds/pairing) between adenine, thymine
and cytosine, guanine;
Name and describe five ways substances can move across the cell-surface
membrane into a cell.
simple diffusion - movement of non polar/small molecules down a concentration gradient
faciliated diffusion - movement of large/polar molecules down a concentration gradient via channel/carrier proteins
osmosis - movement of water down a water potential gradient
Co-transport of 2 different substances using a carrier
protein
Active transport against a concentration gradient via
protein carrier using ATP;
Contrast the structure of the two cells visible in the electron micrographs shown in
Figure 14.
- Magnification (figures) show A is bigger than B;
- A has a nucleus whereas B has free DNA;
- A has mitochondria whereas B does not;
- A has Golgi body/endoplasmic reticulum whereas B
does not; - A has no cell wall whereas B has a murein/glycoprotein
cell wall; - A has no capsule whereas B has a capsule;
- A has DNA is bound to histones/proteins whereas B
has DNA not associated with histones/proteins
OR
A has linear DNA whereas B has circular DNA;
- A has larger ribosomes;
