AS Paper 1 Flashcards
(103 cards)
1
Q
Bacterium vs virus
Is RNA present in b or v
A
Present in bacterium AND virus
2
Q
Bacterium vs virus
Are cell walls present in b or v
A
Present in b not v
3
Q
Bacterium vs virus
R enzyme molecules present in b or v
A
Both present
4
Q
Bacterium vs virus
Is Capsid present in b or v
A
Present in v not b
5
Q
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
A
Complimentary nucleotides pair DNA polymerase
joins nucleotides together to form new phosphodiester bonds
6
Q
Contrast the structures of DNA and mRNA molecule is to give three differences
A
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
7
Q
Describe the difference between the structure of the triglyceride molecule and the structure of a phospholipid molecule
A
in a phospholipd one fatty acid is replaced by a phosphate
8
Q
describe how you would test for the presence of a lipid in a sample of food
A
Add ethanol then add water white emulsion shows lipid
9
Q
Describe have a saturated fatty acid is different from an unsaturated fatty acid
A
Saturated
No double bonds between carbons
10
Q
Figure 1 shows the structure of a fox substitute this fat substitute cannot be digestive in the gods by lipase suggest why
A
The fat substitute is a different shape
So it’s unable to bind to active site of lipase
11
Q
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
A
Its polar
Too large
12
Q
Cells constantly hydrolyse ATP to provide energy describe how ATP is resynthesised in cells
A
From ADP and phosphate
by ATP synthase
during respiration or photosynthesis
13
Q
Give two ways in which the hydrolysis of ATP is used in cells
A
To provide energy for other reactions
to add phosphate to other substances AND make them more reactive
14
Q
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?
A
Can see 3d image
15
Q
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
A
Y is a enzyme
That makes cellulose
16
Q
What is the evidence in figure 3 that the phospholipid bilayer shown is part of the cell surface membrane
A
Cell wall forms outside cell-surface membrane
17
Q
Scientists investigated the hydrolysis of sucrose in growning plant cells by an enzyme called SPS
name the products of the hydrolysis of sucrose
A
Glucose and fructose
18
Q
About the growth of the plant cells from these data? Explain how you reached your conclusions
A
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
19
Q
Describe induced fit model of enzyme action
A
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
20
Q
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
A
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
21
Q
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
A
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
22
Q
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
A
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
23
Q
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
A
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
24
Q
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)
A
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
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
T test
| Comparing two means
26
There is genetic diversity within HPV
| give two ways doctors could use base sequences to compare different types of HPV [2 marks]
You could compare the sequences of DNA or mRNA or look for mutations that change the base sequence
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]
No separation of chromatids chromatids
Chromatids all go to opposite poles
doubles chromosome number in cell
28
Does the mitochondrion have a double outer membrane
Yes
29
Does the chloroplast have a double outer membrane
Yes
30
Does the mitochondrion have starch grains
No
31
Does the chloroplast have starch grains
👍🏼yes
32
Is there diffusion of oxygen into the mitochondrion
Yes
33
Is the diffusion of o2 in the chloroplast
No
34
function of the mitochondria
It’s the site of aerobic respiration or ATP is made
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]
```
```
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;
```
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]
1. 10 cm3 of 10 000 nmol dm–3
/ (original) solution;
2. 90 cm3
of water;
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]
```
Different parts (of amyloid-precursor)
protein;
Each enzyme is specific /fits/binds/
complementary to a different part of the
APP;
```
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.
Peptide bond broken;
| Using water;
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]
```
1. Mutations prevent production of
enzymes
2. (Increase in β-secretase) leads to
more β-amyloid production
3. (Leads to) more/greater plaque
formation;
```
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]
1. (Inhibitor) binds to/blocks active site of
β-secretase/enzyme;
2. Stops/reduces production of βamyloid/plaque;
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]
Some β-amyloid required/needed (to
prevent side effects)
2. Leads to build-up of amyloid-precursor
protein (that causes harm)
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
DNA helicase – (unwinding DNA and) breaking
hydrogen bonds between bases
DNA polymerase – joins (adjacent) nucleotides
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]
```
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;
```
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.:
1. So no contamination/ other bacteria;
2. So same number of bacteria transferred to
allow comparison;
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.
```
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.
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]
1. (Several) values between 10 and 15 (units);
| 2. Repetitions of each;
47
digestion of lipids
(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.
48
digestion of carbs
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.
49
digestion of proteins
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’.
50
Explain the advantages of lipid droplet and micelle formation
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;
51
suggest how the golgi apparatus is involved in the absorption of lipids.
1. Modifies / processes triglycerides;
3. Combines triglycerides with proteins;
4. Packaged for release / exocytosis
OR
Forms vesicles;
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
open close
semi lunar valves 2 3
atrioventicular valves 4 1
53
Use Figure 3 to calculate the volume of blood pumped per minute by the left
ventricle.
```
SV= EDV-ESV
SV = stroke volume
EDV = end-diastolic volume
ESV = end-systolic volume
```
120-40
54
Explain the role of the heart in the formation of tissue fluid.
Contraction of ventricle(s) produces high
blood pressure;
(This) forces water (and some dissolved
substances) out (of blood capillaries);
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]
Excess tissue fluid cannot be (re)absorbed /
| builds up;
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?
buiret
57
Name the chemical element found in all amino acids that is not found in
triglycerides.
n
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
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;
59
Explain how the chromosome number is halved during meiosis.
1. Homologous chromosomes (pair);
2. One of each (pair) goes to each (daughter)
opposite poles;
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?
6
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]
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;
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]
1. homogenise / break tissues
/ cells (in solution);
2. Centrifuge;
3. At increasing speeds
63
give the function of the mitochondrion
The site of aerobic respiration (reactions)
OR
ATP is made
64
feature mitochondria chloroplast
double
outer
membrane
starch grain
diffusion of
o2
in organelles
feature mitochondria chloroplast
double
outer yes yes
membrane
starch grain non yes
diffusion of
o2 yes no
in organelles
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]
Training made no difference to number of
mitochondria per μm
. Training led to an increase in the area (of
inner mitochondrial membrane);
66
Give three properties of water that are important in biology.
[3 marks]
Is a metabolite
Is a solvent
cohesive
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
dilution series
68
Calculate the ratio of final mass to initial mass of potato chips
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
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
(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);
70
Define each of the following terms.
[2 marks]
Species
Species richness
. Species = (A group of) organisms that are
able to produce fertile offspring;
2. Species richness = the number of (different)
species in a community;
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]
. 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;
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).
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);
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).
```
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;
```
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]
(Fewer mitochondria so) less (aerobic)
respiration;
2. (Muscles receive) less ATP (so waste);
75
The renal veins are
The renal veins are blood vessels that return blood to the heart from the kidney
76
Aorta
Renal vein
Vena cava
Aorta to body
Renal vein from kidneys to heart
Vena cava from body to heart
77
Name the blood vessels that carry blood to the heart muscle.
| [1 mark]
Coronary arteries;
78
What is a monomer?
a monomer is a smaller unit from which larger polymers are made;
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
Similarity
1. Both contain galactose / a glycosidic bond;
Difference
2. Lactulose contains fructose, whereas lactose contains glucose;
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.
1. (Lactulose) lowers the water potential of
faeces
2. Water enters (due to osmosis)
and softens the faeces;
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.
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(%);;
82
% decrease =
% decrease = (final-initial) / initial x 100
83
1 Draw box and label a single DNA nucleotide.
1. Phosphate, deoxyribose and base correctly
labelled;
2. Correct shapes and bonds in the correct
positions (as shown below);
84
Give two features of DNA and explain how each one is important in the
semi-conservative replication of DNA.
Weak hydrogen bonds between
bases allow two strands to separate
2. Two strands, so both can act as templates;
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
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.
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.
cell membrane
cell wall
capsule
flagellum
87
2 Name the main biological molecule in:
cytoplasm
cell wall
cytoplasm
phospholipid
cell wall
murein/glycoprotein
88
Name box the process by which prokaryotic cells divide.
binary fission
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.
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
90
Name the type of bond between glycerol and the fatty acids in a triglyceride
ester
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.
(It shows) smaller groups within larger groups /
larger groups containing smaller groups;
2. With no overlap (between groups);
92
Name the taxon to which Drosophilidae belongs
family
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]
Sine song is same length (for
both, so closely related).
2. (But) have different peaks
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.
1. (Three) peaks (in pulse song) occur at the same time (since both female) songs identical
2. (Therefore) no male (song) to cause mating;
95
What is the proteome of a cell?
(The proteome is the full) range of / number of
different proteins that a cell is able to produce (at a
given time);
96
Give two structural differences between a molecule of messenger RNA (mRNA) and
a molecule of transfer RNA (tRNA).
[2 marks]
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;
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]
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);
98
State and explain the property of water that can help to buffer changes in
temperature.
[2 marks]
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.)
99
2 Water is used to hydrolyse ATP.
| Name the two products of ATP hydrolysis.
Adenosine diphosphate and (inorganic)
| phosphate;
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
```
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;
```
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]
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).
102
Explain how one feature of an alveolus allows efficient gas exchange to occur.
[2 marks]
1. (The alveolar epithelium) is one cell thick;
| 2. Creating a short diffusion pathway
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.
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.
