Spec 1 Set 1 Flashcards
Describe how you could use cell fractionation to isolate chloroplasts from leaf tissue.
How to break open cells and remove debris;
Solution is cold/isotonic/buffered;
Second pellet is chloroplast;
Name the parts of the chloroplast labelled A and B.
A stroma;
B granum
Name two structures in a eukaryotic cell that cannot be identified using an optical microscope.
Mitochondria
ribosome
Give one other factor the technician would have controlled.
Concentration of substrate solution
Describe and explain the differences between the two curves at different temperatures.
Initial rate of reaction faster at 37 °C;
Because more kinetic energy;
So more E–S collisions/more E–S complexes formed;
Graph reaches plateau at 37 °C;
Because all substrate used up;
Describe and explain the appearance of one of the chromosomes in cell X
Chromosome is formed of two chromatids;
(Because) DNA replication (has occurred);
(Sister) chromatids held together by centromere;
Describe what has happened during division 1 in Figure 3.
Chromosomes in homologous pair;
One of each into daughter cells / haploid number
Identify one event that occurred during division 2 but not during division 1.
Separation of (sister) chromatids / division of centromere
Name two ways in which meiosis produces genetic variation.
Independent segregation (of homologous chromosomes); Crossing over
During replication, the two strands of a DNA molecule separate and each acts as a template for the production of a new strand.
Enzyme in DNA replication.
DNA Polymerase
The arrows in Figure 5 show the directions in which each new DNA strand is being produced.
Use Figure 4, Figure 5 and your knowledge of enzyme action to explain why the arrows point in opposite directions.
- (Figure 4 shows) DNA has antiparallel strands/described;
- (Figure 4 shows) shape of the nucleotides is different/nucleotides aligned differently;
- Enzymes have active sites with specific shape;
- Only substrates with complementary shape/only the phosphate end (of the developing strand) can bind with active site of enzyme/active site of DNA polymerase;
The scientists obtained DNA from otters that were alive before hunting started.
Suggest one source of this DNA.
their skin
What can you conclude about the effect of hunting on genetic diversity in otters? Use data from Figure 6 to support your answer
Hunting) reduced population size(s), so (much) only few alleles left;
Otters today from one/few surviving population(s);
Some populations of animals that have never been hunted show very low levels of genetic diversity.
Other than hunting, suggest two reasons why populations might show very low levels of genetic diversity.
Due to a genetic bottleneck such as an environmental disaster
Population might have started with small number of individuals
Explain how tissue fluid is formed.
Outward pressure of 3.2 kPa
Forces small molecules out of capillary
The hydrostatic pressure falls from the arteriole end of the capillary to the venule end of the capillary. Explain why.
due to the loss of fluid
High blood pressure leads to an accumulation of tissue fluid. Explain how.
High blood pressure = high hydrostatic pressure;
Increases outward pressure from (arterial) end of capillary/reduces inward pressure at (venule) end of capillary;
(So) more tissue fluid formed /less tissue fluid is reabsorbed;
The water potential of the blood plasma is more negative at the venule end of the capillary than at the arteriole end of the capillary. Explain why.
Water has left the capillary;
Proteins (in blood) too large to leave capillary;
Increasing/giving higher concentration of blood proteins
Name the bond between glycerol and fatty acids in phospholipids.
Ester
The scientists expressed their results as Percentage of lipid in plasma membrane by mass.
Explain how they would find these values.
Divide mass of each lipid by total mass of all lipids (in that type of cell);
Multiply answer by 100;
Suggest one advantage of the different percentage of cholesterol in red blood cells compared with cells lining the ileum.
not supported by other cells so cholesterol helps to maintain shape
E. coli has no cholesterol in its cell-surface membrane. Despite this, the cell maintains a constant shape. Explain why.
Cell unable to change shape;
because it has a cell wall
Explain why the data in Table 3 are described as processed results.
Calculations made (from raw data)
Describe how you would use a 1.0 mol dm–3 solution of sucrose to produce 30 cm3 of a 0.15 mol dm–3 solution of sucrose.
Add 4.5 cm3 of (1.0 mol dm–3) solution
to 25.5 cm3 (distilled) water;
Explain the change in mass of potato tissue in the 0.40 mol dm–3 solution of sucrose.
Water potential of solution is less than/more negative than that of potato tissue;
Tissue loses water by osmosis;
Describe how you would use the student’s results in Table 3 to find the water potential of the potato tissue.
Plot a graph with concentration on the x-axis and percentage change in mass on the y-axis;
Find concentration where curve crosses the x-axis/where percentage change is zero;
Use (another) resource to find water potential of sucrose concentration (where curve crosses x-axis);
HSV infects nerve cells in the face (line 1). Explain why it infects only nerve cells
- Outside of virus has antigens/proteins;
- With complementary shape to receptor/protein in membrane of cells;
- (Receptor/protein) found only on membrane of nerve cells;
HSV can remain inactive inside the body for years (lines 2–3). Explain why this virus can be described as inactive.
- No more (nerve) cells infected/no more cold sores form;
2. (Because) virus is not replicating;
Suggest one advantage of programmed cell death (line 4).
Prevents replication of virus;
Explain how this microRNA allows HSV to remain in the body for years.
MicroRNA binds to cell’s mRNA
- (Binds) by specific base pairing;
- (So) prevents mRNA being read by ribosomes;
- (So) prevents translation/production of proteins;
- (Proteins) that cause cell death;
