Paper 1

Question 1

Which of the follow infectious diseases is caused by bacteria;
Measles
Gonorrhoea
Malaria
HIV

Answer 1

Gonorrhoea

Question 2

Give two symptoms of salmonella poisoning.

Answer 2

Fever, stomach cramps, vomiting , diarrhoea

Question 3

Compare how bacteria and viruses cause symptoms in a host.

Answer 3

Bacteria causes symptoms by producing toxins that damage the host’s cells and tissues.whereas viruses cause symptoms by damaging host cells when they reproduce inside them.

Question 4

Give two features of the bacterium salmonella which shows that it is a prokaryote not eukaryote.

Answer 4

It has got plasmids, single loop of DNA and no nucleus.

Question 5

electron microscope image is produced of a Salmonellabacterium.
* The length of the bacterium in the image is 18 millimetres (mm).
The real length of the bacterium is 4 micrometres (um).
Calculate the magnification of the image.
Use the equation:
magnification = image size/real size

Answer 5

4 micrometre=0.004mm
Magnification=18/0.004=x4500
Or
18mm=18000 micrometre
Magnification=18000/4=x4500

Question 5

Salmonella bacteria can enter a person’s body via contaminated food. They cause illness when they reach the cells of the intestines. Explain how the human body defends itself against infection by salmonella once the pathogen has been ingested.

Answer 5

The stomach produces hydrochloric acid.
The acid helps to kill most of the Salmonella bacteria that reach the stomach from the mouth.
White blood cells/the immune system will try to fight off Salmonella bacteria that reach the cells of the intestines.
For example, some white blood cells will carry out phagocytosis — this involves engulfing the bacteria and digesting them.
Other types of white blood cell will produce antibodies.
These will lock on to the antigens on the Salmonella bacteria and cause them to be targeted for destruction by other white blood cells.
Other types of white blood cell will produce antitoxins.
These will neutralise the toxins produced by the Salmonella.

Question 6

Some pondweed was used to investigate how the amount of light available affects the rate of photosynthesis. The apparatus which was used for this experiment is shown in the figure.
What gas is being collect in the measuring cylinder?

Answer 6

Oxygen

Question 7

What would happen to the volume of gas collected if the investigation was repeated with the lamp off?

Answer 7

The volume of the gas collected will decrease as the intensity of the light has decreased due to the lamp being switched off, so the rate of photosynthesis will decrease too.

Question 8

Sodium hydrogencarbonate dissolves in water and releases carbon dioxide. Suggest why sodium hydrogencarbonate was added to the water in this experiment.

Answer 8

Carbon dioxide is needed for photosynthesis, so adding it to the water ensures that the rate of photosynthesis is not limited by a lack of carbon dioxide.

Question 9

Use the data from the table 1 to calculate the mean rate of photosynthesis over 1 hour. Give your answer in cm3/min. Give your answer to two significant figures.

Answer 9

1 hour= 60 minutes.
Rate of photosynthesis is volume/time or 8/60=0.1333….=0.13cm3/min (2sf)

Question 10

The temperature of the sodium hydrogencarbonate solution was discovered to have increased slightly during the investigation. Explain why this might have affected the rate of photosynthesis of the pondweed?

Answer 10

It may have affected/increased the activity of the enzymes in the pondweed that control photosynthesis.

Increasing the temperature will also have increased the energy of the reacting particles and speed at which they move about (making successful collisions between them more likely). If you’d remembered this bit of particle theory.

Question 11

Suggest how temperature could have controlled in the experiment?

Answer 11

By putting the beaker into a warm water bath to keep the temperature constant.

Question 12

The investigation could have been conducted by counting the number of bubbles given off in a certain amount of time by the pondweed. Suggest one advantage of using a measuring cylinder rather than counting bubbles.

Answer 12

Using a measuring cylinder should give more accurate results.

Question 13

When plants photosynthesise they produce glucose. Give three ways plants use the glucose they produce.

Answer 13

1. For respiration,
2. for making cellulose for cell walls,
3. for making amino acids ( proteins),
4. To store as starch
5. To convert into fats and oils for storage.

Question 14

DIGESTIVE SYSTEM
Describe how starch is broken down in the mouth and small intestine?

Answer 14

Starch is broken down by an enzyme called Amylase into simple sugars like glucose, sucrose, maltose.

Question 15

A sample of food contains starch. The sample is crushed and put into a test tube. A solution containing enzymes is added to the test tube.
Describe a test that could be
Used to determine whether or not the starch
In the sample had been broken down by the enzymes.

Answer 15

1. Add iodine to the solution of the sample, if the iodine solution Remains browny orange then the starch has been broken down. If the solution turns black/blue black then the starch has not been broken down.

Question 16

Bile is a digestive fluid stored in the gall bladder. Which of the following organ produces bile?
Liver, stomach, small intestine, gall bladder.

Answer 16

Liver

Question 17

Gallstones are small, solid stones formed mainly of excess cholesterol. They can block the bile ducts (tubes) that connects the gall bladder to the small intestine.
Explain why eating fatty foods might cause a problem for people suffering from gallstones.

Answer 17

Bile emulsifies fat/breaks fat down into tiny droplets /1 mark).
This gives a larger surface area for enzymes/lipases to work on and so the fat is digested more quickly [1 mark). By blocking the bile ducts, the gallstones could prevent bile from entering the small intestine /1 mark). If so, any fat may be digested more slowly, possibly causing problems.

Question 18

Describe the process by which oxygen moves into cells.

Answer 18

Oxygen moves into cells by diffusion. There’s a net movement of oxygen from higher concentration outside the cells to a lower concentration of oxygen inside the cell through the partially permeable cell membrane.

Question 19

The rate of aerobic respiration increases inside the cell in figure 3. Explain what will happen to the rate of oxygen movement across the cell membrane.

Answer 19

The rate of oxygen movement across the cell membrane will increase. This is becauwill be using more oxygen due to the increased rate of aerobic respiration. This will make the concentration gradient of oxygen across the cell membrane steeper/increase the difference in the concentration of oxygen between the inside and outside of the cell.

Question 20

Trout are fresh water fish. They are relatively large, multicellular organisms with specialised exchange organ. Euglena are small, single celled organisms that live in the water. They don’t have specialised exchange organs. Both Trout and Euglena need oxygen to survive.
Explain why Euglena do not need specialised organs for absorbing oxygen but trout do.

Answer 20

As single-celled organisms, Euglena have a relatively large surface area to volume ratio [1 mark). This means they can absorb enough oxygen to survive by diffusion through their outer surface alone /1 mark). However, as multicellular organisms, trout have a relatively small surface area to volume ratio /1 mark). Diffusion of oxygen across their outer surface would be too slow to supply all their needs, so trout need specialised exchange organs in order to absorb enough oxygen to survive /1 mark].

Question 21

Describe the trend shown in the figure 4

Answer 21

As the rate of work increases the cyclist’s blood lactic acid concentration also increases. At higher rate of work the cyclist’s blood lactic acid concentration increases more quickly.

Question 22

Suggest the reason for the trend you have described above.

Answer 22

During vigorous exercise the body can’t supply enough oxygen to the muscles, so they start to respire anaerobically as well as aerobically /1 mark). Anaerobic respiration produces lactic acid /1 mark). The harder the muscles work, the more they’ll resort to anaerobic respiration and the more lactic acid they’ll produce [1 mark).

Question 23

The cyclist takes part in a sprint race. Explain what will happen to the cyclist’s pulse rate and breathing rate immediately after her race.

Answer 23

The cyclist’s pulse rate and breathing rate will remain high after her race [1 mark). This is because after vigorous exercise the body has an oxygen debt /1 mark). Her pulse rate and breathing rate remain high to supply enough oxygen to react with the build up of lactic acid and so remove it from the cells [1 mark).

Question 24

The cyclist rests after the sprint race. Describe how the cyclist’s heart rate is controlled at rest.

Answer 24

The group of cells in the right atrium of the heart act as a pacemaker.

Question 25

Tobacco smoke contains chemicals that can cause malignant tumours.
Describe why malignant tumours are cancerous but benign tumours are not.

Answer 25

Malignant tumours invade neighbouring tissues / spread to different parts of the body forming secondary tumours (1 mark), whereas benign tumours are contained within one area and do not spread to other parts of the body.

Question 26

Answer 26

Question 27

A scientific magazine used the date in table 2 to report that people who smoke are more likely to die from cancer than who don’t smoke. Does the data support this conclusion.give reasons for your answer.

Answer 27

No. E.g. the data shows the incidence of cancer, not deaths from cancer [1 mark). The data also only shows the incidence of cancer in men not women, so you can’t say this trend is true for everyone [1 mark). Finally the data is only given for smokers, so you can’t compare the likelihood of developing cancer in smokers and non-smokers [1 mark).

Question 28

Calculate the percentage change in height for the plant in beaker C.

Answer 28

% change=(final height-original height)/original height x100
= (7-4)/4x100=75%

Question 29

Explain why the growth in beaker B was poor.

Answer 29

Beaker B was low in nitrates which are needed for making amino acids/ proteins which are essential for growth.

Question 30

Describe how you would expect the pea plant in beaker C which was low in magnesium to look at the end of the week. Explain your answer.

Answer 30

It would have yellow leaves because without magnesium the plant can’t make the chlorophyll that gives it its green colour.

Question 31

State two variables that the students would have had to control for each beaker.

Answer 31

1. The amount of light shining on each beaker
2. The level of substances in the mineral solution.
3. The size of the beakers
   4.the amount of air available
4. The amount of water available.

Question 32

Aside from controlling variables , give one way in which the students could have improved their method in order to obtain more valid results.

Answer 32

They could carry out repeats of the experiment and calculate a mean result for each beaker.

Question 33

Give one conclusion you can draw from this experiment.

Answer 33

Pea plants of this species increase in height at a faster rate with both magnesium and nitrates.

Question 34

Describe what needs to happen to a cell before mitosis can begin.

Answer 34

The cell must grow and increase its number of sub cellular structures.
It must also duplicate its DNA.

Question 35

Describe what will happen next to the cell labelled Y in figure 6

Answer 35

Membranes will form around the two sets of chromosomes / two separate nuclei will form [1 mark]. The cytoplasm and cell membrane will divide [1 mark].
You can see from Figure 6 that the chromosomes in cell Y have been pulled to either end of the cell. That means that the cell is ready to divide into two.

Question 36

A scientist collects 20 cells from a sample.
The cells divide by mitosis once every 20 minutes.
Estimate how many cells will be in the sample after 2 hours.
Give your answer in standard form.

Answer 36

2x60=120 minutes
120/20=6 divisions
So number of cells= 20x26 two power six =1288=1.28x103 10 power 3

If you start with one cell, you need to multiply 2 by itself for the number of divisions (in this case that’s 20). However, as you’re starting with 20 cells here, the total number of cells will be 20 times greater than if you started with just one cell — so you need to multiply 20by 20.

Question 37

Which of the stages A B C shown in figure 7 involve mitosis? Give reason for your answer.

Answer 37

Stage B
This stage results in the division of a single cell into many identical cells/ clones, each of which has a full copy of the genetic material.

Question 38

Read the following information about myeloma.

1. Myeloma is a cancer of the plasma cells — a type of white blood cell.
2. Plasma cells are produced in the bone marrow.
3. Chemotherapy can be used to kill cancerous plasma cells. Chemotherapy is also likely to kill or damage healthy cells in the patient’s bone marrow.
4. A bone marrow transplant can be used following chemotherapy to treat myeloma.
   An individual with myeloma is treated with chemotherapy to kill their cancerous plasma cells. Suggest how a bone marrow transplant from a donor may then be used to treat the individual.

Answer 38

The bone marrow will contain stem cells these will differentiate in the patient to produce new plasma cells replacing those killed during the chemotherapy.

Question 39

Both adult and embryonic stem cells have the potential to be used for medical treatments.
Evaluate the potential use of adult and embryonic stem cells as medical treatments. Your answer which include a justified conclusion.

Answer 39

Adult stem cells are able to differentiate into a smaller range of cell types than embryonic stem cells, which are capable of developing into most types of cell. This means embryonic stem cells may be used to treat a wider range of conditions than adult stem cells.
Adult stem cells which come from the patient’s own body are less likely to be rejected than embryonic stem cells. However, therapeutic cloning offers a way to produce embryonic stem cells that won’t be rejected by the body.
There are ethical issues with using embryonic stem cells, which don’t apply to using adult stem cells. For example, some people feel that human embryos shouldn’t be used as a source of stem cells as each one is a potential human life.

Question 40

Between 1970 and 2000 there was an overall decrease in the number of deaths per year from disease X
Use figure 8 to calculate the mean rate at which the number of deaths per year decreased.

Answer 40

Deaths 1970= 2500
Death in 2000=1000
2500-1000=1500 deaths
2000-1970=30 years
Rate of decrease=15000/30= 50 deaths per year
Read the axes carefully.

Question 41

Use figure 8 to predict the number of deaths from disease X in 2020. Assume the rate of increase from the year 2000 remains constant.

Answer 41

3000

Question 42

Give two steps which scientists carrying out these experiments should have taken in order to prevent the plates being contaminated with unwanted microorganisms.

Answer 42

Any two from: e.g. they should have sterilised the Petri dishes/ culture medium/agar before use. / They should have sterilised the inoculating loops (used to transfer the bacteria to the culture medium) by passing them through a flame. / They should have taped on the lids of the Petri dishes. / They should have stored the Petri dishes/agar plates upside down.
1l

Question 43

Use the experimental evidence from figure 9 to expose the trends shown in the graph in figure 8

Answer 43

Here are some points your answer may include:
The inhibition zones for discs 2 and 3 on the 1975 plate suggest that the antibiotic was very effective against the bacteria that cause disease X — much more effective than the control.
Having an effective antibiotic may have allowed doctors to treat disease X more effectively, therefore reducing the number of deaths from the disease between 1970 and 2000.
However, on the 2005 plate, the inhibition zone is much smaller for disc 3 and non-existent for disc 2, suggesting that the antibiotic is no longer as effective against the bacteria that cause disease X / the bacteria that cause disease X have developed resistance to this antibiotic.
A lack of an effective antibiotic may have caused death rates from disease X to increase between 2000 and 2010, as doctors would have been less able to treat the disease.