AS Paper 1

Question 1

Bacterium vs virus

Is RNA present in b or v

Answer 1

Present in bacterium AND virus

Question 2

Bacterium vs virus

Are cell walls present in b or v

Answer 2

Present in b not v

Question 3

Bacterium vs virus

R enzyme molecules present in b or v

Answer 3

Both present

Question 4

Bacterium vs virus

Is Capsid present in b or v

Answer 4

Present in v not b

Question 5

When HIV infects a human cell, the following events occur,
a single standard length of HIV DNA is made
the human cell then makes a complimentary strand to the HIV DNA
the complimentary strand is made in the same way as a new complimentary strand is made during semiconservative replication of human

DNA describe how the complimentary strand of HIV DNA is made

Answer 5

Complimentary nucleotides pair DNA polymerase

joins nucleotides together to form new phosphodiester bonds

Question 6

Contrast the structures of DNA and mRNA molecule is to give three differences

Answer 6

DNA is double-stranded and mRNA is single-stranded
DNA is very long rna short
T in dna and u in rna
Deoxyribose in dna and ribose in rna

Question 7

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

Answer 7

in a phospholipd one fatty acid is replaced by a phosphate

Question 8

describe how you would test for the presence of a lipid in a sample of food

Answer 8

Add ethanol then add water white emulsion shows lipid

Question 9

Describe have a saturated fatty acid is different from an unsaturated fatty acid

Answer 9

Saturated

No double bonds between carbons

Question 10

Figure 1 shows the structure of a fox substitute this fat substitute cannot be digestive in the gods by lipase suggest why

Answer 10

The fat substitute is a different shape

So it’s unable to bind to active site of lipase

Question 11

This fat substitute is a lipid despite being a lipid it can’t cross the cell surface membrane of cells lining the gut
suggest why it cannot cross cell surface membranes

Answer 11

Its polar

Too large

Question 12

Cells constantly hydrolyse ATP to provide energy describe how ATP is resynthesised in cells

Answer 12

From ADP and phosphate
by ATP synthase
during respiration or photosynthesis

Question 13

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

Answer 13

To provide energy for other reactions

to add phosphate to other substances AND make them more reactive

Question 14

Figure 2 is a photograph of the mitochondrion taken using a scanning electron microscope what’s the evidence from figure 2 that a scanning electron microscope was used to take this photograph?

Answer 14

Can see 3d image

Question 15

Y is a protein
one function of y is to transport cellulose molecules across the phospholipid bilayer
using information from three describe the other function of y

Answer 15

Y is a enzyme

That makes cellulose

Question 16

What is the evidence in figure 3 that the phospholipid bilayer shown is part of the cell surface membrane

Answer 16

Cell wall forms outside cell-surface membrane

Question 17

Scientists investigated the hydrolysis of sucrose in growning plant cells by an enzyme called SPS
name the products of the hydrolysis of sucrose

Answer 17

Glucose and fructose

Question 18

About the growth of the plant cells from these data? Explain how you reached your conclusions

Answer 18

Sucrose hydrolysis linked To some aspect of growth
Greater the rate of hydrolysis more activity as plant grows
Growth remains the same after 8 to 10 days because SPS activity is levelling off

Question 19

Describe induced fit model of enzyme action

Answer 19

The active site isn’t complimentary to the substrate shape of active site changes as substrate binds
Bending bonds in the substrate leading to reaction

Question 20

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 Benedicks test 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 Benedicks test in your answer

Answer 20

Use maltose solutions of known concentrations
Use calorimeter to measure colorimeter value of each solution and plot calibration curve
Find conc of sample from calibration curve

Question 21

HPV is the main cause of cervical cancer.
A vaccine has been developed to protect girls and women from HPV.
describe how giving this vaccine leads to production of antibody against HPV

Answer 21

Vaccine contains antigen from HPV
Displays on antigen-presenting cells 
specific helper T cells detect antigen and stimulate specific B-cell 
B cell divides to give plasma cells
B cell/ plasma cells produces antibody

Question 22

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 Benedicks test 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 Benedicks test in your answer

Answer 22

Use maltose solutions of known concentrations
Use calorimeter to measure colorimeter value of each solution and plot calibration curve
Find conc of sample from calibration curve

Question 23

HPV is the main cause of cervical cancer.
A vaccine has been developed to protect girls and women from HPV.
describe how giving this vaccine leads to production of antibody against HPV

Answer 23

Vaccine contains antigen from HPV
Displays on antigen-presenting cells 
specific helper T cells detect antigen and stimulate specific B-cell 
B cell divides to give plasma cells
B cell/ plasma cells produces antibody

Question 24

Doctors investigated whether it was better to give two or three doses of the HPV vaccine
they determined the main concentration of antibody against HPV in blood samples from girls who were given either two or three doses of the vaccine

girls given two doses received an initial vaccination followed by a second at six months

girls given three doses received an initial vaccination followed by a second at one month and third at six months
the doctors measured the concentration of antibody each month
There is a higher main concentration of antibodies against HPV with the girls that were given two doses of the vaccine compared to girls given three doses
the results are shown in figure 5 what do these results suggest about whether it is better to give two or three doses of the vaccine? Give reasons for your answer.
(2 marks)

Answer 24

Two doses because it has more antibody

with three doses, second dose at one month doesn’t lead to production of any more antibodies

three doses would be more expensive

Question 25

The doctors carried out a statistical test to determine whether the antibody concentrations were significantly different in girls given two doses of the vaccine, compared with those given three doses. They determines the mean concentrations of antibody nine months after the first dose of vaccine
what statistical test should the doctors have you? Give the reason for your choice

Answer 25

T test

Comparing two means

Question 26

There is genetic diversity within HPV

give two ways doctors could use base sequences to compare different types of HPV [2 marks]

Answer 26

You could compare the sequences of DNA or mRNA or look for mutations that change the base sequence

Question 27

Scientists looking for treatments for cancer r investigating the use of substances called Kineson inhibitors
These inhibitors prevent successful mitosis some Kineson inhibitors cause the development of a monopolar spindle in mitosis figure 6 shows chromosomes attached to the normal mitotic spindle and to a monopolar mitotic spindle
suggest why the development of a monopolar mitotic spindle would prevent successful mitosis [2 marks]

Answer 27

No separation of chromatids chromatids

Chromatids all go to opposite poles

doubles chromosome number in cell

Question 28

Does the mitochondrion have a double outer membrane

Answer 28

Yes

Question 29

Does the chloroplast have a double outer membrane

Answer 29

Yes

Question 30

Does the mitochondrion have starch grains

Answer 30

No

Question 31

Does the chloroplast have starch grains

Answer 31

👍🏼yes

Question 32

Is there diffusion of oxygen into the mitochondrion

Answer 32

Yes

Question 33

Is the diffusion of o2 in the chloroplast

Answer 33

No

Question 34

function of the mitochondria

Answer 34

It’s the site of aerobic respiration or ATP is made

Question 35

Scientists investigated the effect of different concentrations of a kinesin inhibitor
(KI) on mitosis of human bone-cancer cells grown in a culture.
Table 3 shows the scientists’ results.
Table 3
Concentration of kinesin inhibitor /
nmol dm–3
Percentage of dividing human
bone-cancer cells showing a
monopolar mitotic spindle
0 0
1 0
10 8
100 93
1000 100
10 000 100
A student who saw these results concluded that in any future trials of this kinesin inhibitor with people, a concentration of 100 nmol dm–3 would be most
appropriate to use.
Do these data support the student’s conclusion? Give reasons for your answer.
[4 marks]

Answer 35

Above 100 may be harmful (to body);
 Higher concentrations more expensive;
above 100) will have more effect on
(rapidly dividing) cancer cells;
 (No, because) at 100 there are still some (7%) cancer cells dividing/undergoing mitosis;

Question 36

At the start of their investigation, the scientists made a solution of kinesin
inhibitor (KI) with a concentration of 10 000 nmol dm–3. They used this to make
the other concentrations by a series of dilutions with water.
Describe how they made 100 cm3 of 1000 nmol dm–3 solution of kinesin inhibitor.
[2 marks]

Answer 36

1. 10 cm3 of 10 000 nmol dm–3
   / (original) solution;
2. 90 cm3
   of water;

Question 37

Read the following passage.
Alzheimer’s disease leads to dementia. This involves small β-amyloid
proteins binding together to form structures called plaques in the brain.
Nerve cells in the brain produce a large protein called amyloid-precursor
protein. This protein is the substrate of two
different enzymes, α-secretase and β-secretase. These enzymes are
normally produced in the brain. One product of the reaction catalysed by
β-secretase is a smaller protein that can lead to β-amyloid protein formation.
Many people with Alzheimer’s disease have mutations that decrease α-secretase production, or increase β-secretase production.
One possible type of drug for treating Alzheimer’s disease is a competitive
inhibitor of β-secretase. When some of these types of drugs were trialled on
patients, the trials had to be stopped because some patients developed
serious side effects.
Use information from the passage and your own knowledge to answer the
following questions.
Suggest how amyloid-precursor protein can be the substrate of two different
enzymes, α-secretase and β-secretase (lines 3–5).
[2 marks]

Answer 37

Different parts (of amyloid-precursor)
protein;
Each enzyme is specific /fits/binds/
complementary to a different part of the
APP;

Question 38

One product of the reaction catalysed by β-secretase is a smaller protein
(lines 6–7).
Describe what happens in the hydrolysis reaction that produces the smaller
protein from amyloid-precursor protein.

Answer 38

Peptide bond broken;

Using water;

Question 39

Many people with Alzheimer’s disease have mutations that decrease
α-secretase production, or increase β-secretase production (lines 8–9).
Use the information provided to explain how these mutations can lead to
Alzheimer’s disease.
[3 marks]

Answer 39

1. Mutations prevent production of
enzymes
2. (Increase in β-secretase) leads to
more β-amyloid production 
3. (Leads to) more/greater plaque
formation;

Question 40

One possible type of drug for treating Alzheimer’s disease is a competitive
inhibitor of β-secretase (lines 10–11).
Explain how this type of drug could prevent Alzheimer’s disease becoming
worse.
[2 marks]

Answer 40

1. (Inhibitor) binds to/blocks active site of
   β-secretase/enzyme;
2. Stops/reduces production of βamyloid/plaque;

Question 41

When some of these types of drugs were trialled on patients, the trials were
stopped because some patients developed serious side effects (lines 11–13).
Using the information provided, suggest why some patients developed serious
side effects.
[1 mark]

Answer 41

Some β-amyloid required/needed (to
prevent side effects)

2. Leads to build-up of amyloid-precursor
   protein (that causes harm)

Question 42

The enzymes DNA helicase and DNA polymerase are involved in DNA
replication.
Describe the function of each of these enzymes.
[2 marks]
DNA helicase
DNA polymerase

Answer 42

DNA helicase – (unwinding DNA and) breaking
hydrogen bonds between bases
DNA polymerase – joins (adjacent) nucleotides

Question 43

Adenosine triphosphate (ATP) is a nucleotide derivative.
Contrast the structures of ATP and a nucleotide found in DNA to give two
differences.
[2 marks]

Answer 43

1. ATP has ribose and DNA nucleotide has
deoxyribose;
2. ATP has 3 phosphate (groups) and DNA
nucleotide has 1 phosphate (group);
3. ATP- base always adenine and in DNA
nucleotide base can be different / varies;

Question 44

A student investigated the effect of three types of disinfectant on the growth of
Lactobacillus bacteria.
During the investigation, the student:
• boiled the agar before pouring the agar plates
• transferred 0.5 cm3 of a diluted liquid culture of Lactobacillus onto each agar
plate
• left some agar plates as controls
• added to other agar plates different concentrations of the disinfectants as
shown in Table 1 on page 5.
After 2 days, she counted the number of colonies of bacteria on each agar plate.
Explain the purpose of:
[2 marks]
boiling the agar:
transferring the same volume of liquid culture onto each agar plate.:

Answer 44

1. So no contamination/ other bacteria;
2. So same number of bacteria transferred to
   allow comparison;

Question 45

A student investigated the effect of three types of disinfectant on the growth of
Lactobacillus bacteria.
During the investigation, the student:
transferred 0.5 cm3 of a diluted liquid culture of Lactobacillus onto each agar
plate

The three disinfectants used by the student were Lysol, propan-2-ol and
ammonia.
Table 1 shows the student’s results.

Concentration of
disinfectant /
arbitrary units
Number of colonies of bacteria
300
The liquid culture the student transferred was diluted by 1 in 10 000 (10–4).
Use information in this question to calculate how many bacteria were present in
1 cm3 of undiluted liquid culture.

Answer 45

As you say, there were 300 bacteria in 0.5 cm3 (?) of the 10000-fold diluted solution - think this way: if half a cm3 (which 0.5 cm3 is, agreed?) contains 300 bacteria, then surely a whole cm3 i.e. (double of the 0.5 cm3) will contain twice as many bacteria, correct? Equals 600 bacteria in one cubic centimetre [one cm3].

But you were also told that this 0.5cm3 of solution was 10000 times weaker than the original i.e. that it had been diluted by a factor of 10000.

SO DO YOU AGREE THAT the original solution (BEFORE DILUTION) would have had many more bacteria? Of course, well done!

NEXT Q: How many more? YES, 10000 times as many!.

So your answer is: 600 X 10000 = 6000000

= 6 X 10E6 [the figure 6 followed by 6 zeros] which is the same as 6 million bacteria/cm3.

Question 46

Concentration of disinfectant / arbitrary units :
Number of colonies of bacteria

    Lysol Propan-2-ol Ammonia
0  300 300 300
5  0 290 300
10 0 195 295
15 0 0 275
20 0 0 240 

The student concluded that the minimum concentration of propan-2-ol needed to
stop the growth of Lactobacillus was 15 units.

This conclusion is incorrect.
Describe how you could obtain a more accurate estimate of the minimum
concentration of propan-2-ol needed to stop the growth of this species of
bacterium.
[2 marks]

Answer 46

1. (Several) values between 10 and 15 (units);

2. Repetitions of each;

Question 47

digestion of lipids

Answer 47

(at the lumen of SI)

Lipids: Bile emulsifies fats in the stomach for digestion to happen faster. It is secreted by the liver. Pancreatic lipase hydrolyses the fats into monoglycerides and fatty acids. Monoglycerides and fatty acids combine with bile salts and phospholipids to form micelles. The bile slats and phospholipids allows the poorly soluble monoglycerides and fatty acids to the cells lining the intestinal walls where the non-polar nature of these molecules help them to diffuse through the lipid bi-layer.

Question 48

digestion of carbs

Answer 48

Carbohydrates: Amylase in the saliva hydrolyses starch into two maltose molecules. Pancreatic juices that are secreted into the stomach contain amylase too. Maltase is embedded in the intestinal wall which hydrolyses the maltose into two glucose molecules. This is then absorbed into the blood stream by co-transport which is explained in ‘Biological molecules AQA AS Biology PART 8 of 8 TOPICS: Inorganic ions’. NB: It is important that you say hydrolyses and not break down as you will not get the mark.

Question 49

digestion of proteins

Answer 49

Proteins: Amino peptidase from the pancreas is an exopeptidase where it hydrolyses the ends of the polypeptide to make amino acids. Trypsin is an endopeptidase where it hydrolyses the proteins into polypeptides in the middle. Cells from the intestinal wall secrete enzymes called dipeptidase which hydrolyse dipeptides into amino acids. This is then absorbed into the blood using co-transport which is similar to glucose and is further explained in ‘Biological molecules AQA AS Biology PART 8 of 8 TOPICS: Inorganic ions’.

Question 50

Explain the advantages of lipid droplet and micelle formation

Answer 50

1. Droplets increase surface areas (for lipase /
   enzyme action);
2. (So) faster hydrolysis / digestion (of
   triglycerides / lipids);
3. Micelles carry fatty acids and glycerol /
   monoglycerides to / through membrane / to
   (intestinal epithelial) cell;

Question 51

suggest how the golgi apparatus is involved in the absorption of lipids.

Answer 51

1. Modifies / processes triglycerides;
2. Combines triglycerides with proteins;
3. Packaged for release / exocytosis
   OR
   Forms vesicles;

Question 52

Figure 3 shows the volume changes in the left ventricle of a human heart during
two cardiac cycles. The numbers 1, 2, 3 and 4 represent times when heart valves
open or close.

Use information from Figure 3 to complete Table 2. Place the number 1, 2, 3 or
4 in the appropriate box.

                                      open                   close semi lunar valves atrioventicular valves

Answer 52

open close
semi lunar valves 2 3
atrioventicular valves 4 1

Question 53

Use Figure 3 to calculate the volume of blood pumped per minute by the left
ventricle.

Answer 53

SV= EDV-ESV
SV	=	stroke volume
EDV	=	end-diastolic volume
ESV	=	end-systolic volume

120-40

Question 54

Explain the role of the heart in the formation of tissue fluid.

Answer 54

Contraction of ventricle(s) produces high
blood pressure;
(This) forces water (and some dissolved
substances) out (of blood capillaries);

Question 55

Lymphoedema is a swelling in the legs which may be caused by a blockage in
the lymphatic system.
Suggest how a blockage in the lymphatic system could cause lymphoedema.
[1 mark]

Answer 55

Excess tissue fluid cannot be (re)absorbed /

builds up;

Question 56

Scientists measured the mean amino acid concentration in white wines made from
grapes grown organically and white wines made from grapes that were not grown
organically.
0 5 . 1 Which test could the scientists have used to identify that there are amino acids in
white wine?

Answer 56

buiret

Question 57

Name the chemical element found in all amino acids that is not found in
triglycerides.

Answer 57

n

Question 58

The scientists used a statistical test to determine whether there was a significant
difference in the amino acid concentration in the two types of white wine.
They obtained a value for P of 0.04.
Name the statistical test the scientists used and give a reason for your answer.
Was the difference significant? Give a reason for your answer.
[3 marks]

Name of statistical test

Reason for choice

Explanation of test result

Answer 58

Choice: (Student’s) t-test;
Reason for choice: Looking for differences
between two means;
Explanation: Difference is significant / not due to
chance because the P value is less than
0.05;

Question 59

Explain how the chromosome number is halved during meiosis.

Answer 59

1. Homologous chromosomes (pair);
2. One of each (pair) goes to each (daughter)
   opposite poles;

Question 60

Figure 6 shows a cell from the moss plant.
The cell is in the second meiotic division.
Figure 6
0 6 . 3 What is the haploid number of chromosomes for this species of moss?

Answer 60

6

Question 61

Crossing over greatly increases genetic diversity in this species of moss.
Describe the process of crossing over and explain how it increases genetic
diversity.
[4 marks]

Answer 61

1. Homologous pairs of chromosomes associate
   / form a bivalent
2. Chiasma(ta) form;
3. (Equal) lengths of (non-sister) chromatids /
   alleles are exchanged;
4. Producing new combinations of alleles;

Question 62

Describe how you would use cell fractionation techniques to obtain a sample of
chloroplasts from leaf tissue. Do not include in your answer information about
any solutions.
[3 marks]

Answer 62

1. homogenise / break tissues
   / cells (in solution);
2. Centrifuge;
3. At increasing speeds

Question 63

give the function of the mitochondrion

Answer 63

The site of aerobic respiration (reactions)
OR
ATP is made

Question 64

feature mitochondria chloroplast

double
outer
membrane

starch grain

diffusion of
o2
in organelles

Answer 64

feature mitochondria chloroplast

double
outer yes yes
membrane

starch grain non yes

diffusion of
o2 yes no
in organelles

Question 65

Scientists investigated the effect of an exercise programme on the number and
size of mitochondria in skeletal muscle. They took samples of muscle from a
large number of volunteers before and after the exercise programme. From each
sample, they cut thin sections and used these to determine the mean number of
mitochondria per μm2 and the mean area of inner mitochondrial membranes.
Their results are shown in Figure 7 and Figure 8.
Figure 7 Figure 8
What do the data in Figure 7 and Figure 8 suggest about the effect of the
exercise programme on mitochondria?
[2 marks]

Answer 65

Training made no difference to number of
mitochondria per μm

. Training led to an increase in the area (of
inner mitochondrial membrane);

Question 66

Give three properties of water that are important in biology.
[3 marks]

Answer 66

Is a metabolite
Is a solvent
cohesive

Question 67

The student produced the sucrose solutions with different concentrations from a
concentrated sucrose solution.
Name the method she would have used to produce these sucrose solutions.
[1 mark]
Name of method

Answer 67

dilution series

Question 68

Calculate the ratio of final mass to initial mass of potato chips

Answer 68

You express them as a single number because they are all x : 1

E.g. for the first one it would be 3.82/2.79 : 2.79/2.79 which is 1.37:1, so you just write 1.37

Question 69

Concentration of
sucrose

solution / mol dm–3
Initial mass of
chip / g

Final mass of
chip / g

0.0 2.79 3.82
0.2 2.75 2.97
0.4 2.78 2.67
0.6 2.69 2.31
0.8 2.72 2.20
1.0 2.77 1.99
Explain the result for the chip in 0.8 mol dm–3 sucrose solution

Answer 69

(0.8 mol dm–3
sucrose) solution has a more
negative water potential than potato
(cytoplasm);
2. (therefore) water moves out (of potato) into
the (sucrose) solution by osmosis (so cells
decrease in mass);

Question 70

Define each of the following terms.
[2 marks]
Species
Species richness

Answer 70

. Species = (A group of) organisms that are
able to produce fertile offspring;
2. Species richness = the number of (different)
species in a community;

Question 71

Scientists investigated the species richness of fish caught at various depths in the
Pacific Ocean close to the western coast of Chile.
Figure 9 shows the scientists’ results. 68% of all the fish caught in this
investigation came from sample A
68% of all the fish caught in this investigation came from sample A.
A student thought this showed that sample A had a greater index of diversity than
any of the other samples.
It is not possible to draw this conclusion from the given data. Give reasons why.
[3 marks]

Answer 71

. Number of individuals of each species not
known;
2. Almost all (of sample A / the 68%) could be of
the same species;
3. Two / other samples have a higher number of
species / higher species richness but a lower
number of individuals / fish;
4. Other samples may have more individuals of
each species;

Question 72

0 Read the following passage.
Azidothymidine (AZT) is a drug used to treat people infected with human
immunodeficiency virus (HIV). It inhibits the enzyme that synthesises
DNA from HIV RNA. This does not destroy HIV in the body but stops or
slows the development of AIDS.

In the past, some people who took AZT on its own eventually developed
AIDS. Some of the HIV in their bodies had become resistant to AZT.
To prevent this from happening, people infected with HIV are now treated
with highly active antiretroviral therapy (HAART). This involves taking AZT
with other anti-HIV drugs at the same time.
AZT is taken in low doses. This is because people who took high doses
over long periods of time suffered muscle wastage. It was found that high
doses of AZT inhibit replication of mitochondria.
Use information from the passage and your own knowledge to answer the
questions.
5
10
1 0 . 1 Suggest and explain why AZT does not destroy HIV in the body but stops or
slows the development of AIDS (lines 3–4).

Answer 72

1. Person (infected with HIV) has HIV DNA (in
   their DNA);
2. New HIV (particles) still made;
3. (AZT) inhibits reverse transcriptase;
4. (AZT) stops these (new HIV particles) from
   forming new HIV DNA;
5. Stops destruction of more / newly infected
   T cells;
6. So immune system continues to work (and
   AIDS does not develop);

Question 73

0 Read the following passage.
Azidothymidine (AZT) is a drug used to treat people infected with human
immunodeficiency virus (HIV). It inhibits the enzyme that synthesises
DNA from HIV RNA. This does not destroy HIV in the body but stops or
slows the development of AIDS.

In the past, some people who took AZT on its own eventually developed
AIDS. Some of the HIV in their bodies had become resistant to AZT.
To prevent this from happening, people infected with HIV are now treated
with highly active antiretroviral therapy (HAART). This involves taking AZT
with other anti-HIV drugs at the same time.
AZT is taken in low doses. This is because people who took high doses
over long periods of time suffered muscle wastage. It was found that high
doses of AZT inhibit replication of mitochondria.
Use information from the passage and your own knowledge to answer the
questions.
5Suggest and explain two advantages of using HAART (lines 7–9).

Answer 73

Slows / stops the development of AIDS;
2. Because HIV resistant to AZT is damaged from replicating (by
other drugs);
OR
3. AZT continues to work as a drug;
4. Because HAART prevents the spread of
AZT-resistant HIV to rest of the human
population so No new HIV particles made;
6. Because HAART might interfere with viral
protein synthesis;

Question 74

0 Read the following passage.
Azidothymidine (AZT) is a drug used to treat people infected with human
immunodeficiency virus (HIV). It inhibits the enzyme that synthesises
DNA from HIV RNA. This does not destroy HIV in the body but stops or
slows the development of AIDS.

In the past, some people who took AZT on its own eventually developed
AIDS. Some of the HIV in their bodies had become resistant to AZT.
To prevent this from happening, people infected with HIV are now treated
with highly active antiretroviral therapy (HAART). This involves taking AZT
with other anti-HIV drugs at the same time.
AZT is taken in low doses. This is because people who took high doses
over long periods of time suffered muscle wastage. It was found that high
doses of AZT inhibit replication of mitochondria.
Use information from the passage and your own knowledge to answer the
questions.
Suggest why high doses of AZT lead to muscle wastage (lines 10–11).
[2 marks]

Answer 74

(Fewer mitochondria so) less (aerobic)
respiration;
2. (Muscles receive) less ATP (so waste);

Question 75

The renal veins are

Answer 75

The renal veins are blood vessels that return blood to the heart from the kidney

Question 76

Aorta
Renal vein
Vena cava

Answer 76

Aorta to body
Renal vein from kidneys to heart
Vena cava from body to heart

Question 77

Name the blood vessels that carry blood to the heart muscle.

[1 mark]

Answer 77

Coronary arteries;

Question 78

What is a monomer?

Answer 78

a monomer is a smaller unit from which larger polymers are made;

Question 79

Lactulose is a disaccharide formed from one molecule of galactose and one molecule
of fructose.
Other than both being disaccharides, give one similarity and one difference between
the structures of lactulose and lactose.
[2 marks]
Similarity
Difference

Answer 79

Similarity
1. Both contain galactose / a glycosidic bond;
Difference
2. Lactulose contains fructose, whereas lactose contains glucose;

Question 80

Following digestion and absorption of food, the undigested remains are processed to form faeces in the parts of the intestine below the ileum.
The faeces of people with constipation are dry and hard. Constipation can be treated
by drinking lactulose. Lactulose is soluble, but is not digested or absorbed in the
human intestine.
Use your knowledge of water potential to suggest why lactulose can be used to help
people suffering from constipation.

Answer 80

1. (Lactulose) lowers the water potential of
   faeces
2. Water enters (due to osmosis)
   and softens the faeces;

Question 81

Lactulose can also be used to treat people who have too high a concentration of
hydrogen ions (H+) in their blood.
The normal range for blood H+ concentration is 3.55 × 10−8 to 4.47 × 10−8 mol dm−3

A patient was found to have a blood H+ concentration of 2.82 × 10−7 mol dm−3

Calculate the minimum percentage decrease required to bring the patient’s blood H+ concentration into the normal range.

Answer 81

you need the % reduction from 2.82 X 10-7 to 4.47 X 10-8 (this latter one is the higher of the normal range so will give the minimum value needed).

First, make the exponent the same in both:

So we need %age reduction from 28.2 X 10-8 to 4.47 X 10-8

You can now ignore the exponents:-
SO: Reduction needed = 2.82 × 10-7 − 4.47 × 10-8 /
2.82 ×10-7

(-) 84.1(%);;

Question 82

% decrease =

Answer 82

% decrease = (final-initial) / initial x 100

Question 83

1 Draw box and label a single DNA nucleotide.

Answer 83

1. Phosphate, deoxyribose and base correctly
   labelled;
2. Correct shapes and bonds in the correct
   positions (as shown below);

Question 84

Give two features of DNA and explain how each one is important in the
semi-conservative replication of DNA.

Answer 84

Weak hydrogen bonds between
bases allow two strands to separate
2. Two strands, so both can act as templates;

Question 85

Replication of mitochondrial DNA (mtDNA) is different from that of nuclear DNA. box
The replication of the second strand of mtDNA only starts after two-thirds of the first
strand of mtDNA has been copied.
A piece of mtDNA is 16 500 base pairs long and is replicated at a rate of 50
nucleotides per second.
Tick ( ) the box that shows how long it would take to copy this mtDNA.
A 330 seconds
B 440 seconds
C 550 seconds
D 660 seconds

Answer 85

16500 base pairs long and you need to copy both strands.

The first strand starts copying at 50nt per second and reaches 2/3 of it’s length (i.e. 10999nt) in 219.98 seconds (i.e. 10999/50).
(2/3 of 16500 is 11000)

At this point second strand begins and needs to finish all 16500nt at 50nt per second (16500/50) =330 seconds.

Whole thing would be finished in 219.98+330=550 seconds.

Question 86

Figure box 3 shows part of a prokaryotic cell.
Figure 3
0 4 . 1 Name the structures labelled W to Z in Figure 3.

Answer 86

cell membrane
cell wall
capsule
flagellum

Question 87

2 Name the main biological molecule in:
cytoplasm
cell wall

Answer 87

cytoplasm
phospholipid

cell wall
murein/glycoprotein

Question 88

Name box the process by which prokaryotic cells divide.

Answer 88

binary fission

Question 89

Some prokaryotic cells can divide every 30 minutes. A liquid culture contained a
starting population of 1.35 × 10^4 cells.
Assuming each cell divides every 30 minutes, calculate how many cells there will be
after 3 hours. Assume no cells die during this time.

Answer 89

if it divides every 30 mins, that means the population doubles every 30 mins.
3 hours = 6 x 30 mins, so it doubles 6 times, or 2^6. Multiply the original population by 2^6

8.64 x 105

Question 90

Name the type of bond between glycerol and the fatty acids in a triglyceride

Answer 90

ester

Question 91

Figure box 5 represents the phylogenetic classification of four different species of fruit fly.
Figure 5
0 6 . 1 Figure 5 shows a hierarchy. Explain how.

Answer 91

(It shows) smaller groups within larger groups /
larger groups containing smaller groups;
2. With no overlap (between groups);

Question 92

Name the taxon to which Drosophilidae belongs

Answer 92

family

Question 93

Drosophila box fruit flies display courtship behaviour. One of the stages of courtship is
singing by males. Normally a male will produce a ‘sine song’, in which continual noise
is made, and a ‘pulse song’, in which there is continual noise with some louder peaks.
Scientists showed fruit flies a visual stimulus that made them sing. They made
recordings of these songs.
Figure 6 shows the recordings of the songs of three flies over the same time period.

D. erecta and D. willistoni are closely related species but different species.
Describe evidence from Figure 6 that supports this statement.
[2 marks]

Answer 93

Sine song is same length (for
both, so closely related).
2. (But) have different peaks

Question 94

The scientists repeated their experiments, using female fruit flies as the visual
stimulus. When a male and female D. willistoni were together, their songs led to
mating.
When two female D.willistoni were together, their songs did not lead to any attempt to
mate.
Use information from Figure 6 to suggest why the two females did not attempt to
mate.

Answer 94

1. (Three) peaks (in pulse song) occur at the same time (since both female) songs identical
2. (Therefore) no male (song) to cause mating;

Question 95

What is the proteome of a cell?

Answer 95

(The proteome is the full) range of / number of
different proteins that a cell is able to produce (at a
given time);

Question 96

Give two structural differences between a molecule of messenger RNA (mRNA) and
a molecule of transfer RNA (tRNA).
[2 marks]

Answer 96

mRNA has codons, tRNA has an anticodon;

mRNA is linear / straight chain, tRNA is
cloverleaf;

2. mRNA does not have an amino acid binding site, 0tRNA does;
3. mRNA has more nucleotides;
4. (Different) mRNAs have different lengths, all tRNAs are similar / same length;

Question 97

Starting with mRNA in the cytoplasm, describe how translation leads to the production of a polypeptide.
Do not include descriptions of transcription and splicing in your answer.
[5 marks]

Answer 97

1. ribosome attaches to mRNA;
2. Ribosome moves to the start codon /
   AUG;
3. tRNA brings specific amino acid;
4. Anticodon (on tRNA complementary) to codon (on mRNA);
5. Ribosome moves along to next codon;
6. (Process repeated and) amino acids join by peptide bonds / condensation reaction (to form polypeptide);

Question 98

State and explain the property of water that can help to buffer changes in
temperature.
[2 marks]

Answer 98

1. (water has a relatively) high (specific) heat
   capacity;
2. Can gain / lose a lot of heat / energy
   without changing temperature;
   (Hydrogen bonds between water molecules can absorb a lot of energy – This means that water has a high specific heat capacity - it takes a lot of energy to heat it up.)

Question 99

2 Water is used to hydrolyse ATP.

Name the two products of ATP hydrolysis.

Answer 99

Adenosine diphosphate and (inorganic)

phosphate;

Question 100

Hydrolysis of ATP is catalysed by the enzyme ATP hydrolase.
A student investigated the effect of ATP concentration on the activity of ATP
hydrolase. She used shortening of strips of muscle tissue caused by contraction as
evidence that ATP was being hydrolysed.
• She took four slides A, B, C and D, and added strips of muscle tissue of the same
length to each slide.
• She then added the same volume of ATP solutions of different concentrations to the
four slides and left each slide for five minutes.
• She then recorded the final length of each strip of muscle tissue.
Her results can be seen in Table 1.
Table 1
Slide : Concentration of ATP solution added to slide / ×10 6 mol dm 3 : Final length of muscle tissue after 5 minutes / mm
A : 2 : 36
B : 4 : 31
C : 6 : 29
D : 8 : 26
Other than those given, name two variables the student should have controlled.
Describe box and explain the pattern shown by the data in Table 1.
[2 marks]
Description
Explanation

Answer 100

1. Species  the muscle tissue
came from;
Temperature of the muscle tissue / ATP
solution / slides;
3. pH of the ATP solution;

Description
1. As concentration of ATP increases, length
of muscle decreases;
Explanation
2. More ATP (hydrolysed by ATP
hydrolase), so more energy released, so
more muscle contraction / shortening of
muscle;

Question 101

The hydrolysis of 1 dm3 of a 1 mol dm−3 solution of ATP releases 30 500 J of energy.
60% of the energy released during the hydrolysis of 1 mol dm−3 of ATP is released as heat; the rest is used for muscle contraction.
The student added 0.05 cm3 of ATP solution to slide D.
Calculate the energy available from ATP for contraction of the muscle on this slide.
[3 marks]

Answer 101

the hydrolysis of 1 dm3 of a 1moldm-3 solution releases 30500j of energy…………….” Assuming this solution (in bold) is a solution containing ATP, then let us go step by step, OK?

30500 joules of energy corresponds to one mole of ATP (there is one mole of ATP in 1 dm3 of a 1M (1 Molar) solution of ATP (info given).

So how much is 40% (not lost as heat) of this? Of course, 30500 X 0.4 = 12200 joules is given by one mole of ATP, yeah?

OK next step:-
They tell you that 0.05 cm3 of soln was used - so how much ATP in it?
Yes do it bit by bit……………… well done! 0.05 cm3 = 0.05/1000 dm3 = 5 X 10-5 dm3
Conc-n given = 8 X 10-6 M
So, actual MOLES of ATP in it = (8 X 10-6) X (5 X 10-5) = 40 X 10-11 = 4 X 10-10 moles

Last step:
FROM italics above,
one mole of ATP gives 12200 joules of energy
So how much will underlined amount give?

YES: 4 X 10-10 X 12200 = 12.2 X 4 X 10-7 = 48.8 X 10-7 joules = 4.88 X 10-6 joules (ANSWER).

Question 102

Explain how one feature of an alveolus allows efficient gas exchange to occur.
[2 marks]

Answer 102

1. (The alveolar epithelium) is one cell thick;

2. Creating a short diffusion pathway

Question 103

Carbon monoxide is a poisonous gas that is present in cigarette smoke. This carbon monoxide can be absorbed into the blood where it binds with haemoglobin.
Scientists investigated the concentration of carbon monoxide in cars in which people
were smoking or not smoking. They measured the concentration with the car
windows open and closed. Figure 7 shows the scientists’ results as they presented
them. A value of ± 2 standard deviations from the mean includes over 95% of the
data.

In England, in October 2015, a law was introduced making it illegal to smoke in a car
carrying someone who is under the age of 18.
Following the introduction of the law, a politician stated:
‘It is dangerous to smoke when a child is in the car. Higher levels of deadly toxins can
build up, even on short journeys, and children breathe faster than adults, meaning
they inhale more of the deadly toxins.’
Use the information provided and the data in Figure 7 to evaluate the politician’s
statements.

Answer 103

Carbon monoxide is a poisonous gas that is present in cigarette smoke. This carbon monoxide can be absorbed into the blood where it binds with haemoglobin.
Scientists investigated the concentration of carbon monoxide in cars in which people
were smoking or not smoking. They measured the concentration with the car
windows open and closed. Figure 7 shows the scientists’ results as they presented
them. A value of ± 2 standard deviations from the mean includes over 95% of the
data.

In England, in October 2015, a law was introduced making it illegal to smoke in a car
carrying someone who is under the age of 18.
Following the introduction of the law, a politician stated:
‘It is dangerous to smoke when a child is in the car. Higher levels of deadly toxins can
build up, even on short journeys, and children breathe faster than adults, meaning
they inhale more of the deadly toxins.’
Use the information provided and the data in Figure 7 to evaluate the politician’s
statements.