AQA BIO OLD

Question 1

When HIV infects a human cell, the following events occur.
A single-stranded length of HIV DNA is made.
The human cell then makes a complementary strand to the HIV DNA.
The complementary strand is made in the same way as a new complementary
strand is made during semi-conservative replication of human DNA.
Describe how the complementary strand of HIV DNA is made.

Answer 1

Nucleotides are attracted to the exposed bases by complementary base pairing. DNA Polymerase catalyses a condensation reaction which forms phosphodiester bonds between adjacent new nucleotides forming a new new polynucleotide strand

Question 2

Contrast the structures of DNA and mRNA molecules to give three differences.

Answer 2

1. DNA is made of deoxyribose sugar whereas mRNA is made of ribose sugar
2. DNA is double stranded whereas mRNA is single stranded
3. DNA has hydrogen bonding whereas mRNA does not.
4. DNA contains the base Thymine whereas mRNA contains the base Uracil

Question 3

Describe the difference between the structure of a triglyceride molecule and the structure of a phospholipid molecule.

Answer 3

Triglyceride is made of 3 fatty acids and one glycerol whereas a Phospholipid molecule is made of one phosphate and two fatty acids and a glycerol.

Question 4

Figure 1 shows the structure of a fat substitute.
This fat substitute cannot be digested in the gut by lipase
Suggest why.

Answer 4

Fat substitute is not a complementary shape to the shape of lipase enzyme active site so unable to bind to active site of lipase so an enzyme substrate complex is not formed so a reaction cannot be catalysed.

Question 5

Cells constantly hydrolyse ATP to provide energy.

Describe how ATP is resynthesised in cells.

Answer 5

ATP is resynthesised in a condensation reaction between ADP and Pi catalysed by ATP synthase.

Question 6

Give two ways in which the hydrolysis of ATP is used in cells,

Answer 6

1. To provide energy for other reactions

2. To add phosphate to other substances and make them more reactive

Question 7

What is the evidence from Figure 2 that a scanning electron microscope was
used to take this photograph?

Answer 7

A 3D image can be seen in figure 2

Question 8

Y is a protein. One function of Y is to transport cellulose molecules across the
phospholipid bilayer.
Using information from Figure 3, describe the other function of Y.

Answer 8

Y is an enzyme that makes cellulose by joining beta glucose molecules together

Question 9

Describe the induced-fit model of enzyme action.

Answer 9

Before the reaction the active site of the enzyme is not complementary to the substrate, the shape of the active site changes as the substrate binds which distorts certain bonds in the substrate leading to a reaction

Question 10

Explain the results shown. (Graph plateaus after initial high rate)

Answer 10

The rate of increase in concentration of maltose slows as starch is used up and there is little increase after 25 minutes because there is little starch left

Question 11

A quantitative Benedict’s test produces a colour whose intensity depends on the
concentration of reducing sugar in a solution. A colorimeter can be used to
measure the intensity of this colour.
The scientist used quantitative Benedict’s tests to produce a calibration curve of colorimeter reading against concentration of maltose.
Describe how the scientist would have produced the calibration curve and used
it to obtain the results in Figure 4.
Do not include details of how to perform a Benedict’s test in your answer.

Answer 11

Make maltose solution of different concentrations ie. from 0.0 to 1.0 increasing by 0.2 mol/dm3 and carry out the test on each concentration. Use a colorimeter to measure colorimeter value of each solution and plot calibration curve described in the question. Find concentration of the sample from calibration curve.

Question 12

Suggest why the development of a monopolar mitotic spindle would prevent successful mitosis.

Answer 12

Sister chromatids are not separated which means sister chromatids are not pulled to opposite ends of the cell so one daughter cell will have no chromosomes and one daughter cell will have double the number of chromosomes.

Question 13

A student who saw these results concluded that in any future trials of this kinesin inhibitor with people, a concentration of 100 mol dm would be most appropriate to use.

Do these data support the student’s conclusion? Give reasons for your answer.

Answer 13

No because at 100 there are still 7% of cancel cells dividing so cancer is not destroyed and may continue to grow and form tumours. The ideal concentration may be between 100 and 100 but there is no data for this so further trials are needed for concentrations between 100 and 100. However above 100 may be harmful to the body as side effects have not been recorded and higher concentrations will be more expensive.

Question 14

Suggest an advantage of using a pH meter rather than a pH indicator

Answer 14

Gives quantitative data values rather than subjective qualitative data values which are less accurate.

Question 15

Explain why the pH decreases when the lipase is added to the milk

Answer 15

The lipids in the milk are hydrolysed into fatty acids plus glycerol which decreases the pH since it’s an acid

Question 16

Suggest why the pH remained constant after two minutes

Answer 16

All the triglycerides have been used up and converted into fatty acids and glycerol so no more fatty acids produced

Question 17

Name the part of the pancreas that produces the inactive form of trypsin

Answer 17

Ribosome

Question 18

Suggest the advantage of producing trypsin in an inactive form inside the cells in the pancreas

Answer 18

Does not digest protein inside the cells so pancreas tissues is not damaged

Question 19

Explain how a competitive inhibitor stops an enzyme such as trypsin from working

Answer 19

The competitive inhibitor has a complementary shape to the active site of trypsin, the same shape as trypsin’s substrate. The inhibitor will bind to the active site of trypsin forming an enzyme inhibitor complex so the trypsin can no longer bind to its substrate and catalyse its hydrolysis, and fewer enzyme substrate complexes are formed so it cannot work.

Question 20

Explain the role of the diaphragm when breathing out

Answer 20

Intercostal muscles relax so ribcage is pulled in and down. The diaphragm also relaxes and domes up which decreases the volume in the thorax so pressure increases. Pulmonary pressure is higher than atmospheric pressure so air moves down the concentration gradient out of the lungs.

Question 21

Use the data shown to compare the change in FEV1 of people who continued to sample with those who stopped smoking

Answer 21

The FEV1 of people who stopped smoking decreases from 2.78 dm3 to 2.71 dm3 whereas the FEV1 of people who continued to smoke decreases by a lot more from 2.78 dm3 to 2.48 dm3. So the effect on the lungs was much greater as the rate of decrease was higher.

Question 22

Smoking causes changes in the lungs and airways of smokers. Suggest two chnages in the lungs of people who continued to smoke that could explain the change in their FEV1.

Answer 22

1. Airways are narrowed

2. Scar tissues builds up

Question 23

Suggest how milk can be treated to remove lactose

Answer 23

Add lactase which is the enzyme which catalyses the hydrolysis of lactose

Question 24

The scientists told the volunteers to drink the milk first thing in the morning rather than at bedtime. Suggest why

Answer 24

Would be able to record symptoms if they are awake not if they r tired or asleep

Question 25

Suggest one instruction that the scientists would have given about what they shouldn’t eat or drink that day during this investigation

Answer 25

Eat no other foods containing lactose

Question 26

Suggest why the scientists changed the type of milk they have each group after one week

Answer 26

Each group acts as it’s own control so the effect of lactose can be compared between both groups

Question 27

People who do not have the specific receptor protein in their cell surface membrane may be infected with the Ebola virus but do not develop the disease. Explain why they do not develop the disease.

Answer 27

The virus cannot bind to these specific receptor proteins in the cell surface membrane so cannot enter the cell and replicate to produce more virus particles and so doesn’t damage cells

Question 28

Explain the increase in specific plasma cells and antibody in people infected with the Ebola virus

Answer 28

Immune system recognises virus as foreign and T- lymphocytes stimulate B-lymphocytes to divide rapidly by mitosis into plasma cells which release antibodies which bind to and immobilise/ kill the virus particles

Question 29

Explain how a blood transfusion from a patient recently recovered from Ebola may be an effective treatment

Answer 29

Lots of antibodies in recovered patient and the transfusion contains antibodies so the antibodies which are specific to the Ebola virus particles while bind with the antigens on the Ebola virus particles and the virus is destroyed/ cannot enter the cell

Question 30

A high mutation rate makes it difficult to develop a vaccine. Explain why.

Answer 30

The virus repeatedly changes its protein coat so the antigens of the virus are different due to high antigenic variability . If the mutation rate is high a vaccine won’t be effective as the vaccine contains specific antigen so the antibodies not complementary to changed antigens

Question 31

Glucose is absorbed from the lumen of the small intestine into epithelial cells Explain how the transport of sodium ions in involved in absorption of glucose by epithelial cells.

Answer 31

Sodium ions is actively transported out of epithelial cells in the small intestine into the blood stream through carrier proteins. This means there’s is a higher concentration of sodium ions outside the epithelial cells than inside the cells so sodium ions move by facilitated diffusion down the concentration gradient into the epithelial cells. Glucose is cotransported with sodium into the epithelial cells through sodium-glucose cotransporter proteins.

Question 32

Explain why the diffusion of chloride ions involves a membrane protein and the diffusion of oxygen does not.

Answer 32

Chloride ions are charged and polar so cannot cross the phospholipid bilayer membrane so are transported by facilitated diffusion through carrier proteins. Oxygen however is not charged and is non polar so can diffuse across the phospholipid bilayer membrane.

Question 33

Use your knowledge of water potential to explain why adults who cannot digest lactose get diahorrea when they drink milk

Answer 33

Lactose remains in the intestine so there is a lower water potential in the intestine so water enters by osmosis from body cells into the intestine

Question 34

The equation for breakdown of lactose. Name the type of chemical reaction

Answer 34

Lactose+ Lactase —> Glucose+ Galactose Hydrolysis

Question 35

Describe how you could use the Biuret test to distinguish a solution of the enzyme lactase from a solution of lactose. Why would you expect this result with lactase

Answer 35

Add Biuret’s Reagent to both solutions. The solution containing lactase should turn from blue to purple/ violet as it is a protein.

Question 36

Describe the path by which oxygen goes from an alveolus to the blood

Answer 36

Oxygen dissolved in the inner moist lining of the alveolus and diffuses through the alveolar epithelium through the capillary endothelium into the blood

Question 37

Explain why people with miner lungs have a lower concentration of oxygen in their blood

Answer 37

The alveolar epithelium becomes thicker so there is an increased diffusion distance for oxygen so less oxygen reaches the blood

Question 38

Describe how ventilation helps to maintain a difference in oxygen concentration between the alveoli and oxygen.

Answer 38

Ventilation removes air with a lower oxygen concentration and brings in air with a higher oxygen concentration

Question 39

Give one other way that helps to maintain the difference in oxygen concentration

Answer 39

Circulation of the blood

Question 40

Coal mining in Britain had been dramatically reduced by 1990. Some scientists concluded that the rise in reported cases of miners lung after 1992 shows that the disease takes a long time to develop.

Answer 40

The graph shows fluctuations and correlation does not mean causation as there may be other causes of miners lungs. Furthermore there will be improved diagnosis methods post 2000 compared to before the year 2000 and not all cases may have been reported earlier than this as not all individuals with the condition visit a doctor.

Question 41

Describe what the graph shows about the effect on substrate concentration on the rate of this enzyme controlled reaction

Answer 41

The rate of reaction increases directly proportionally to substrate concentration until 23 units where the graph plateaus and remains at a steady rate

Question 42

What limits the rate of reaction between points A and B ( the start of the curve vs where it begins to plateau)

Answer 42

The substrate concentration limits the rate of reaction as substrate concentration increases the rate increases directly proportionally to this.

Question 43

Suggest a reason for the shape between point C and D (where the rate remains constant after plateauing)

Answer 43

All actives sizes of enzymes are occupied so the maximum number of E-S complexes has been formed

Question 44

Explain how the competitive inhibitor drug lowers the rate of reaction controlled by folate reductase

Answer 44

The methotrexate has a shape which is similar to the substrate of folate reductase so is complementary in shape to the active site of folate reductase and binds to it forming a enzyme inhibitor complex so substrate can no longer bind so less E-S complexes are formed.

Question 45

Methotrexate only affects the rate of reaction of one enzyme, why doesn’t it affect other enzymes

Answer 45

It’s shape is only complementary to the shape of the active site of folate reductase as it is similar in shape to the enzymes specific substrate

Question 46

There are a large number of mitochondria in this cell, explain how these organelles help the cell to absorb the products of digestion.

Answer 46

Mitochondria are the site of aerobic respiration to release energy/ produce ATP. ATP is required for the active transport as it occurs against the concentration gradient and is an active process.

Question 47

Explain why people with coeliac disease can digest proteins only have small concentrations of amino acids in the blood

Answer 47

Villi are damaged so there is a decrease in surface area of villi so less absorp

Question 48

What is the advantage of giving results as a ratio for changes in mass?

Answer 48

It allows for comparisons as the discs have different starting masses

Question 49

The student were advised that they could improve the relativity of their results by taking additional readings at the same concentration of sodium chloride

Answer 49

Allows for a mean to be calculated which reduces the effect of anomalies and anomalies can also be identified

Question 50

The students used a graph of their results to find the sodium chloride solution with the same water potential as the apple tissue

Answer 50

Plot points of ratio of mass at the start to mass at end on the y-axis against the concentration of sodium chloride solution on the x-axis and draw a line of best fit. Find the point where the ratio is one where the line of best fit crosses the x axis.

Question 51

The student were advised they could improve thier growth by taking additional readings. Explain how.

Answer 51

Line of best fit is more precise so point at which the line crosses the x-axis is more precise so more accurate conclusion can be made

Question 52

Explain why monoclonal antibodies are referred to as monoclonal.

Answer 52

They are produced from the same B cell which was cloned

Question 53

Give two ways in which a pathogen may cause disease when it has entered the body

Answer 53

1. Produces and releases toxins | 2. Damages healthy cells

Question 54

Why is a buffered, ice cold isotonic solution used for cell fractionation?

Answer 54

Buffered: Maintains the pH to prevent proteins from denaturing Ice Cold: stops enzyme activity which may damage organelles Isotonic: prevent movement of water into or out of the cell causing cell lysis

Question 55

Tests using monoclonal antibodies are specific. Use your knowledge of protein structure to explain why.

Answer 55

Each protein has a specific primary structure or order or amino acids which means each protein has a specific tertiary 3D structure so each protein is only complementary in shape to one antigen

Question 56

The tests using monoclonal antibodies allow vets to identify brucellosis while they are still on the farm. Explain the advantages of this.

Answer 56

Infected case can be isolated quickly and culled so infected diary products are not sold which reduces the spread of disease so non infected animals do not need to be killed

Question 57

The cardiac cycle is controlled by the sinonatrial node and the atrioventricular node. Describe how.

Answer 57

SAN initiates heartbeat and sends waves of electrical impulses across the walls of the atria causing atrial contraction. The waves reach the Annulus Fibrosis which is a region of non-conducting tissue which prevents the waves spreading straight to the ventricles. AVN delays electrical impulses which allows for the atria to empty before the ventricles contract. The AVN send waves of electrical activity along the BUNDLE OF HIS ( collection of conducting tissue) which divides into the Purkyne Fibres which are two conducting fibres and spread around the ventricles and initiate the ventricles to contract from the base of the heart

Question 58

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

Answer 58

Add leaf to isotonic/ cold and buffered solution. Homogenise leaf tissue using homogeniser/ blender. Filter to remove unbroken leaf tissue and cellular debris. Centrifuge at a low speed for a short period of time to separate and collect pellet containing nuclei. Pour supernatant into second tube and centrifuge at a higher speed for a longer time period to separate and collect pellet containing chloroplasts.

Question 59

Describe and explain the difference between the two curves at 25 and 37 degrees

Answer 59

The initial rate of reaction is faster at 37 degrees as particles gain more kinetic energy so more enzyme substrate complexes are formed. The graph plateaus at 37 degrees because all of the substrate has been converted into products.

Question 60

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

Answer 60

Chromosome is composed of two chromatids because DNA replication has occurred with sister chromatids held together by centromere

Question 61

What happens during division one of meiosis?

Answer 61

Homologous chromosomes pulled to opposite sides of the cells, centromeres remain intact as sister chromatids are not separate so one of each into daughter cells

Question 62

Identify one event that occurs during division 2 but not during division 1

Answer 62

Centromere is divided as sister chromatids separated and pulled to opposite poles of the cell

Question 63

Name two ways in which meiosis produced genetic variation

Answer 63

1. Independent segregation | 2. Crossing over of allleles

Question 64

The arrows show the directions in which each new DNA strand is being produced. Use your knowledge of enzyme action to explain why the arrow point in different directions.

Answer 64

DNA strands are antiparrallel so one runs from the 5 to the 3 while the other runs from the 3 to the 5. Enzyme has active site with a specific shape which is complementary to its substrate. Only substrates with a complementary shape can bind with active site of the enzyme

Question 65

Use the information in the figure to explain how tissue fluid is formed

Answer 65

Hydrostatic pressure is greater than osmotic pressure by 3.2 Kpa so fluid moves down the pressure gradient out of the capillary by ultrafiltration through fenestrations in the cell wall

Question 66

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

Answer 66

Loss of fluid at arteriole end but larger proteins remain in the artery as they cannot pass through fenestration in capillary wall causing hydrostatic pressure to fall as greater concentration of water outside of the cell

Question 67

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

Answer 67

High blood pressure means higher hydrostatic pressure at the arteriole end of the capillary so more fluid leaves the capillary so more tissue fluid is formed. Excess of tissue fluid so lymphatic system is unable to drain the excess tissue fluid leading to accumulation.

Question 68

The water potential if the blood plasma is more negative at the venule end of the capillary than at the arteriole end of the capillary. Explain why.

Answer 68

Loss of water at the artiole end but larger plasma proteins remain in the blood as they are too large to leave the capillary leading to a higher concentration of blood proteins at the venule end decreasing the water potential of the blood plasma.

Question 69

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

Answer 69

Red blood cells are free in the blood so are not supported by other cells so cholesterol helps to maintain the shape

Question 70

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

Answer 70

Cell is unable to change shape because it has a cell membrane all which is rigid and made of murein.

Question 71

HSV only affects nerve cells. Explain why it only affects nerve cells

Answer 71

Antigens found on the surface of the virus are specifically shaped and complementary to active site of receptors found on cell membrane of cell. Only receptors on nerve cells are complementary to antigen is virus.

Question 72

HSV can remind inactive in the body for years. Explain why this virus can be described as inactive.

Answer 72

RNA not converted into DNA so no more nerve cells are infected as the virus is not spreading

Question 73

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

Answer 73

MicroRNA binds to cell’s mRNA by specific base pairing which prevents mRNA being read by ribosomes so prevents production of proteins that cause cell death.

Question 74

Describe and explain how centrifuging the culture allowed the scientists to obtain a cell free liquid

Answer 74

Centrifuge the sample at a high speed causing more dense heavy material to be collected in the form of a pellet at the bottom of the test tube. Filter to separate supernatant from pellet.

Question 75

Give one function of lysosomes

Answer 75

Break down of cell organelles/ waste

Question 76

Suggest one advantage of H pylori cells being able to produce enzyme that neutralises acid

Answer 76

Prevent them from being damaged by stomach acid which is highly acididc

Question 77

What does the data suggest about the damage caused to human cells by the toxin and by the enzyme that neutralises acid?

Answer 77

There is more cell damage when both toxin and acid neutralising enzyme are present. Some cell damage when toxin or acid neutralising enzyme are on their own as seen in B and C. The stand deviation does not overlap with A with B and C so there is a statistically significant difference. Standard deviation overlap between B and C so there is not a real difference between the isolated toxin or enzyme.