2019 P1 Flashcards
A student concluded from Figure 3 that eating an extra 10g of fibre per day would significantly lower his risk of cardiovascular disease. Evaluate (3)
.NEGATIVE CORRELATION between fibre eaten per day and risk of cardiovascular disease
.Original fibre intake of student is not known
.Correlation doesn’t mean causation
.NO STATS TEST so can’t be concluded differences are significant
Suggest one advantage and one disadvantage of using the FFQ method compared with the alternative method (2)
Advantage - Over longer period so more representative
Disadvantages - Relies on long term memory so may not be accurate
Use Figure 5 to explain how human mass at birth is affected by stabilising selection (3)
.Most likely to be transferred to a special care unit are those under 2800g
.Extreme mass babies least likely to survive AND so less likely to pass on their alleles
.Extreme mass at birth decreases in frequency
FILL IN (3)
- allele
- locus
- transcribed
- translated
- golgi
- tertiary
Tick one box that gives the name of the statistical test that the scientists should use with the data in Table 2 to test this null hypothesis (1)
- Chi-squared
P value 0.03. What can you conclude? (3)
probability that difference in frequency of births above 4500g is due to chance is LESS THAN 0.05
so REJECT null hypothesis
presence of KIR2DSI does significantly affect frequency of high birth mass
Describe structure of HIV
Genetic material is RNA Reverse transcriptase Protein capsomere Phospholipid envelope Attachment proteins
Tick one box that shows the number of virus particles that would be present in a test sample of blood taken from an HIV controller with the median viral load.
- 106
Use the data in Table 3 and your knowledge of the immune response to suggest why HIV controllers do not develop symptoms of AIDS. (3)
.All have more T helper cells
.LowER viral load TO infect help T cells
.so activation of B cells
.with B cells more production of plasma cells
Describe and explain the data in Table 4 (2)
cell cycle of mice heart cells
Trend of slowing growth from before birth to 21 days
DNA replication happens before mitosis
Describe how BrdU would be incorporated into new DNA during semi-conservative replication (5)
DNA helicase
breaks hydrogen bonds between 2 DNA strands
BrdU complementary to ADENINE on template strand
DNA polymerase joins NUCLEOTIDES to incorporate BrdU into the new DNA strand
PHOSPHODIESTER bonds form between nucleotides
Use your knowledge of the ELISA test to suggest and explain how the scientists identified the cells that have BrdU in their DNA (3)
Add antibody to cells
Wash cells to remove excess antibody
Add substrate to cause colour change
Complete Table 5 (making 30cm3 of this sodium chloride solution) (2)
cm3 volume of water/cm3
0.8 19.2
Use data in Table 6 to calculate concentration of sodium chloride solution with a water potential of -3.41MPa (2)
0.07 mol dm-3
Use Figure 8 and Figure 9 to calculate by how much this leaf increased in area (2)
9 cm3
Suggest and explain one way the leaf growth of xerophytic plants would be different from the leaf growth of sunflowers in Figure 9 (2)
slow growth
due to smaller number of stomata for gas exchange
Use you knowledge of gas exchange in leaves to explain why plants grown in soil with very little water grow slowly (2)
stomata close
so less carbon dioxide uptake for less photosynthesis
Use this equation to calculate how many time faster the metabolic rate of a mouse is than the metabolic rate of a horse (2)
15 times faster
Suggest how these differences allow the mouse to have a higher metabolic rate than the horse (2)
Mouse HAEMOGLOBIN has lower affinity for oxygen
MORE oxygen can be dissociated for metabolic reactions
Use your knowledge of surface area to volume ratio to explain the higher metabolic rate of a mouse compared to a horse (3)
Mouse has larger surface area to volume ratio as it is smaller
Mouse has more heat loss
Faster rate of respiration releases heat
