June 2022 Component 2 Past Paper Flashcards
(31 cards)
What is meant by polymorphism (1)
. Many alleles
. For same game
. Cannot be maintained by mutation alone
State the null hypothesis (1)
. No statistical difference between the observed and expected results
Explain how the vegetation of the soil type at site A and B shown in image 1.2 could have caused the difference in banding of the snail at the 2 sites (3)
. Vegetation acts as a selection pressure
. Site A short grass and dark soil more bands and site B fewer bands provide better camouflage
. Better camoflaged increased chance of survival
. So can reproduce and pass on advantageous alleles
Suggest how the location and habitat of the sites has led to significant changes in frequency of the allele between the 2 population of this species of snail (3)
. Sea acts a geographical barrier between the population / allopathic speciation
. Different selection pressures
. Different alleles for random mutation will be different in each population
. Genetic drift due to no interbreeding
Describe the production of amylase was triggered in the barley seeds
. Absorption of water
. Gibberlins released by embryo/ hydrolysed
. Caused release of amino acids from aleurone layer
. Amino acid used to synthesise amylase
Explain why it was important that the same concentration of agar and the same concentration of starch was used in all Petri dish (2)
. Agar no effect on rate of diffusion
. Starch same number of molecules to digest
. Both are controlled variable
Explain why range bars are drawn on a graph (2)
. From maximum to minimum
. Increased reliability
Suggest a reason for the decrease in amylase activity after 8 days of germination (3)
. Leaves appear at 8 days
. Which is able to photosynthesis
. So plant can produce it own glucose
. Therefore less starch needed to be broken down
Identify A-G (3)
A
A
F
C
Name the type of epithelium found in both B and the trachea and describe the function of one tissue in both location (3)
. Cillated
. B = moves blastocyst/ fertilised ovum
. Trachea move mucus out of breathing system
Explain why the cell harvested from the grafian follicle is a secondary occyte and not a female gamete (2)
. Has not completed meiosis / meiosis 2 only completed on entry of sperm
. Chromosome still have 2 chromatids / has not released second polar body
Describe how the entry of more than one sperm into the secondary oocyte would be prevented (2)
. Cortical reaction releases content of cortical granules
. Zona pellicuda thicken
Suggest why the type of IVF shown in image 3.2B has a higher rate of success in producing an embryo (2)
. In B sperm does not have to digest a path to secondary oocyte / does not need acrosome reaction
. In B nucleus injected directly into 2nd oocyte
. A may no be able to penetrate the zona pellicuda
Explain the advantage to the zygote of this unequal division of cytoplasm during the formation of the secondary oocyte (2)
. Zygote contains nearly all of the cytoplasm
. So can divide rapidly with less time needed between cell division
Name the tissue in plant where these specialised protein molecule would be found and explain why the solution used in the hydroponic systems needs to be oxygenated (3)
. Endodermis
. Cadmium have to be actively transported into cells
. Synthesis of ATP
State the type of variation in all 3 population (1)
. Continuous
Exaplin why it would be inappropriate to use either a chi test a student t test to determine the significant difference between the survival of seedlings grown from seeds collected from A and B (2)
. Cannot use chi2 because data is continuous
. Cannot use t test because data is not normally distributed
Scientists conducted a long term project to reduce cadmium concentration to safe level in a habitat with a cadmium concentration of 15.au.. evaluate decision (4)
. At high conc only C can survive
. But C can only survive at Cd above 10 Au
. So would die out as Cd falls
. So need to grow B then A
Because B would die out when Cd falls below 2au
Type of reaction and bond (2)
Condensation reaction
Ester
Glycosidic
Peptide
Deduce the identify of these bases giving reason for your answer (3)
Base X = guanine
Base Y = cytosine
. X is a double ring and Y is a single ring
. Cannot be adenine and thymine as T not found in RNA
Explain why the addition of dideoxyribonucleotide to end of a DNA strand prevent further bases being added to the chain (1)
. DNA polymerase active site is not complementary
Explain why the fragments have seperated as shown (3)
. Fragments are negatively charged
. Some move towards the positive electrode
. Smaller fragments move faster
. Because they can move through the pore in the gel more easily
State one medical advantage of using NGS rather than Sanger sequencing (1)
. Faster diagnosis
. Treatment can be tailored to individual patients
Find epu
28 epu