Spec 1 Set 1

Question 1

Describe how you could use cell fractionation to isolate chloroplasts from leaf tissue.

Answer 1

How to break open cells and remove debris;
Solution is cold/isotonic/buffered;
Second pellet is chloroplast;

Question 2

Name the parts of the chloroplast labelled A and B.

Answer 2

A stroma;

B granum

Question 3

Name two structures in a eukaryotic cell that cannot be identified using an optical microscope.

Answer 3

Mitochondria

ribosome

Question 4

Give one other factor the technician would have controlled.

Answer 4

Concentration of substrate solution

Question 5

Describe and explain the differences between the two curves at different temperatures.

Answer 5

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;

Question 6

Describe and explain the appearance of one of the chromosomes in cell X

Answer 6

Chromosome is formed of two chromatids;
(Because) DNA replication (has occurred);
(Sister) chromatids held together by centromere;

Question 7

Describe what has happened during division 1 in Figure 3.

Answer 7

Chromosomes in homologous pair;

One of each into daughter cells / haploid number

Question 8

Identify one event that occurred during division 2 but not during division 1.

Answer 8

Separation of (sister) chromatids / division of centromere

Question 9

Name two ways in which meiosis produces genetic variation.

Answer 9

Independent segregation (of homologous chromosomes);
Crossing over

Question 10

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.

Answer 10

DNA Polymerase

Question 11

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.

Answer 11

1. (Figure 4 shows) DNA has antiparallel strands/described;
2. (Figure 4 shows) shape of the nucleotides is different/nucleotides aligned differently;
3. Enzymes have active sites with specific shape;
4. 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;

Question 12

The scientists obtained DNA from otters that were alive before hunting started.
Suggest one source of this DNA.

Answer 12

their skin

Question 13

What can you conclude about the effect of hunting on genetic diversity in otters? Use data from Figure 6 to support your answer

Answer 13

Hunting) reduced population size(s), so (much) only few alleles left;
Otters today from one/few surviving population(s);

Question 14

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.

Answer 14

Due to a genetic bottleneck such as an environmental disaster
Population might have started with small number of individuals

Question 15

Explain how tissue fluid is formed.

Answer 15

Outward pressure of 3.2 kPa

Forces small molecules out of capillary

Question 16

The hydrostatic pressure falls from the arteriole end of the capillary to the venule end of the capillary. Explain why.

Answer 16

due to the loss of fluid

Question 17

High blood pressure leads to an accumulation of tissue fluid. Explain how.

Answer 17

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;

Question 18

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.

Answer 18

Water has left the capillary;
Proteins (in blood) too large to leave capillary;
Increasing/giving higher concentration of blood proteins

Question 19

Name the bond between glycerol and fatty acids in phospholipids.

Answer 19

Ester

Question 20

The scientists expressed their results as Percentage of lipid in plasma membrane by mass.
Explain how they would find these values.

Answer 20

Divide mass of each lipid by total mass of all lipids (in that type of cell);
Multiply answer by 100;

Question 21

Suggest one advantage of the different percentage of cholesterol in red blood cells compared with cells lining the ileum.

Answer 21

not supported by other cells so cholesterol helps to maintain shape

Question 22

E. coli has no cholesterol in its cell-surface membrane. Despite this, the cell maintains a constant shape. Explain why.

Answer 22

Cell unable to change shape;

because it has a cell wall

Question 23

Explain why the data in Table 3 are described as processed results.

Answer 23

Calculations made (from raw data)

Question 24

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.

Answer 24

Add 4.5 cm3 of (1.0 mol dm–3) solution

to 25.5 cm3 (distilled) water;

Question 25

Explain the change in mass of potato tissue in the 0.40 mol dm–3 solution of sucrose.

Answer 25

Water potential of solution is less than/more negative than that of potato tissue;
Tissue loses water by osmosis;

Question 26

Describe how you would use the student’s results in Table 3 to find the water potential of the potato tissue.

Answer 26

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);

Question 27

HSV infects nerve cells in the face (line 1). Explain why it infects only nerve cells

Answer 27

1. Outside of virus has antigens/proteins;
2. With complementary shape to receptor/protein in membrane of cells;
3. (Receptor/protein) found only on membrane of nerve cells;

Question 28

HSV can remain inactive inside the body for years (lines 2–3). Explain why this virus can be described as inactive.

Answer 28

1. No more (nerve) cells infected/no more cold sores form;

2. (Because) virus is not replicating;

Question 29

Suggest one advantage of programmed cell death (line 4).

Answer 29

Prevents replication of virus;

Question 30

Explain how this microRNA allows HSV to remain in the body for years.

Answer 30

MicroRNA binds to cell’s mRNA

1. (Binds) by specific base pairing;
2. (So) prevents mRNA being read by ribosomes;
3. (So) prevents translation/production of proteins;
4. (Proteins) that cause cell death;