2.4- cell recognition and immune system Flashcards
7 box . 1 When a person is bitten by a venomous snake, the snake injects a toxin into the
person. Antivenom is injected as treatment. Antivenom contains antibodies against
the snake toxin. This treatment is an example of passive immunity.
Explain how the treatment with antivenom works and why it is essential to use passive
immunity, rather than active immunity.
[2 marks]
- (Antivenom/Passive immunity) antibodies
bind to the toxin/venom/antigen and
(causes) its destruction; - Active immunity would be too slow/slower;
1. For ‘bind’ accept ‘attach’, ignore ‘attack’. 1. For ‘destruction of toxin’ accept agglutination or phagocytosis. 1. Ignore reference to antibodies ‘neutralising toxin/stopping damage’ 1. Reject reference to ‘killing’ toxin/venom. 2. Accept ‘passive immunity is faster’, not simply ‘passive immunity is fast’.
This question was all based on section 3.2.4 of the specification, in the novel context of production
of snake antivenom.
There were two parts to question 07.1 – how does an antibody work and what is the difference
between passive and active immunity? 28.1% of students could answer both of these correctly.
Confusion was demonstrated between antigens on a toxin and antibody binding to a pathogen.
Errors resulting in the first marking point not being awarded included describing binding, but failing
to discuss destruction, or discussing destruction without reference to binding. Commonly, students
used the idea of ‘complementary’ in place of binding. Those who were not awarded marking point
two tended not to make a comparative statement about active and passive immunity, and some
made vague statements about active immunity ‘taking time’, rather than expressing length of time.
A mark of zero was commonly the result of confusing active and passive immunity or treating the
antivenom as a vaccine which would then trigger an immune response.
Figure 8 shows a procedure used to produce antivenom. box
Figure 8
7 . 2 A mixture of venoms from several snakes of the same species is used.
Suggest why.
[2 marks]
. May be different form of antigen/toxin (within one species) OR Snakes (within one species) may have different mutations/alleles; 2. Different antibodies (needed in the antivenom) OR (Several) antibodies complementary (to several antigens);
In 07.2, many students gave answers relating to not needing to identify the species of snake that
had bitten a person, as the antivenom would work against the venom of several species – this was
not creditworthy as the question is clearly related to several snakes of the same species. A pleasing number of students understood the production of antivenom and appreciated the need for
several antibodies to be produced by the animal to be used in the patient. Some suggested that
one antibody could be effective against several antigens; this was not given credit. Pleasingly, few
students confused antigens and antibodies. Common errors included referring to different forms of
venom, which was indicated in the question stem, or failing to include reference to antibodies and
instead making vague statements about “being effective against” or “fighting off” or
“neutralising/counteracting” the venom. Roughly even proportions of students scored two, one and
zero marks for this question
0 7 box . 3 Horses or rabbits can be used to produce antivenoms.
When taking blood to extract antibody, 13 cm3 of blood is collected per kg of the
animal’s body mass.
The mean mass of the horses used is 350 kg and the mean mass of the rabbits used
is 2 kg
Using only this information, suggest which animal would be better for the production of
antivenoms.
Use a calculation to support your answer.
[2 marks]
- Horses because more
antivenom/antibodies could be collected
(as more blood collected);
2. 4550 (cm3 ) v 26 (cm3 ) (blood collected) 2. Accept 175 rabbits needed to (collect the volume of blood from) one horse.
In question 07.3, most students could complete the required calculation. Even so, many students
suggested the rabbit would be better as it would likely be safer for the rabbit as less blood was
removed or suggested that the same number of antibodies would be produced in a smaller volume
of blood. Of those who did pick the horse, many only stated that more blood could be collected,
rather than linking this to more antibodies/antivenom being collected from each animal. 81% of
students scored at least one mark here.
0 7 . 4 During the procedure shown in Figure 8 the animals are under ongoing observation
by a vet.
Suggest one reason why [1 mark]
- (So) the animal does not suffer from the
venom/vaccine/toxin; - (So) the animal does not suffer
anaemia/does not suffer as a result of
blood collection; - (So) the animal does not have pathogen
that could be transferred to humans;
Too many generalised answers were given to question 07.4. At this level, a specific reason why it
would be ethical to have veterinary supervision in this particular procedure was required.
Confusion was sometimes demonstrated here over whether the animals were being administered
venom or antivenom.
Accept ‘To fulfil licence/legal requirements’. Accept ‘(So) the animal does not have pathogen that could result in it producing other antibodies (not wanted in the antivenom)’. For ‘pathogen’ accept correct form of pathogen.
7 box . 5 During vaccination, each animal is initially injected with a small volume of venom.
Two weeks later, it is injected with a larger volume of venom.
Use your knowledge of the humoral immune response to explain this vaccination
programme.
[3 marks]
- B cells specific to the venom reproduce by
mitosis; - (B cells produce) plasma cells and memory
cells; - The second dose produces antibodies (in
secondary immune response) in higher
concentration and quickly
OR
The first dose must be small so the animal
is not killed;
1. Accept in context of primary or secondary immune response. 1. Credit idea of specificity if given once in relation to T or B cell. 1. Accept a description for specificity. 1. Accept ‘clone’ for ‘reproduce by mitosis’. 1. ‘Clonal selection of B cells’ = MP1. 3. Accept ‘a lot of antibody’ for ‘higher concentration of antibody
In 07.5, very few students could give a complete account of the humoral immune response in this
context, but those who could (7.3%) gave some excellent answers. It was rare to see the idea of
specificity to the venom antigens being key to the B-cells cloning, and confusion was demonstrated
between T-cells and B-cells and which produced antibodies. About half of the students achieved
the second marking point for recognising the two differentiated forms of B-cells, but beyond this
many did not achieve any further marks due to vague statements or misconceptions. A small
number of students demonstrated a more detailed understanding but still did not obtain marks due
to key ideas being missed. Examples included the idea of specificity of T/B-cells missing from their
answer for marking point 1, mitosis/cloning missing from marking point 2, or one of the ideas of
high concentration or ‘quickly’ missing from marking point 3.
0 7 . 1 Describe how phagocytosis of a virus leads to presentation of its antigens.
[3 marks]
- Phagosome/vesicle fuses with lysosome;
- (Virus) destroyed by lysozymes/hydrolytic
enzymes; - Peptides/antigen (from virus) are displayed on
the cell membrane; - Accept vacuole
fuses with lysosome - Reject virus fuses
with lysosome
07.1 Many students did not refer to the antigen being presented on the surface membrane of the
phagocyte, so could not be awarded mark point 3.
0 7 . 2 Describe how presentation of a virus antigen leads to the secretion of an antibody
against this virus antigen.
[3 marks]
- Helper T cell/TH cell binds to the antigen (on
the antigen-presenting cell/phagocyte); - This helper T/TH cell stimulates a specific B
cell; - B cell clones
OR
B cell divides by mitosis; - (Forms) plasma cells that release antibodies;
07.2 Some students described, once again, how the virus would be presented. Mark points 3
and 4 were more often awarded than mark points 1 and 2, suggesting better understanding
of the actions of B cells in the immune response than T cell involvement. Mark point 1
ould be awarded if the student stated that a T cell binds to the antigen and then
differentiates into a T helper cell.
0 7 . 3 Collagen is a protein produced by cells in joints, such as the knee.
Rheumatoid arthritis (RA) is an auto-immune disease. In an auto-immune disease,
a person’s immune system attacks their own cells. RA causes pain, swelling and
stiffness in the joints.
Scientists have found a virus that produces a protein very similar to human
collagen.
Suggest how the immune response to this viral protein can result in the
development of RA.
[2 marks]
- The antibody against virus (antigen) will bind
to collagen; - This results in the destruction of the (human)
cells/collagen;
07.3 The responses to this question revealed much misunderstanding of the immune response
as a whole, with many references to ‘thinking’ immune systems. Many students simply
repeated phrases from the question stem; for example, “since the virus protein and the
human collagen have a similar shape, the immune system will attack the human collagen”.
Students needed to identify that the part of the immune response which would ‘attack’ the
collagen would be the binding of specific antibodies, and then to use their knowledge of
how an antigen-antibody complex leads to the destruction of the antigen (section 3.2.4 of
the specification), i.e., human collagen in this case. Credit could be gained for reference to
agglutination or phagocytosis as methods of ‘attacking’ the human collagen, since these
are the methods of antigen destruction named in the specification.
0 1 . 2 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.
[3 marks]
- (Complementary) nucleotides/bases pair
OR
A to T and C to G; - DNA polymerase;
- Nucleotides join together (to form new
strand)/phosphodiester bonds form; - & 3. Ignore ‘(DNA
polymerase) forms
base pairs/nucleotide
pairs’
If clearly writing rote
answer about DNA
replication 2 max e.g.
helicase or separating
strands
Question 1.2 discriminated very well, with 15% obtaining three marks and 21% scoring zero.
There were good, concise answers that scored three marks for including complementary base
pairing and the role of DNA polymerase in joining nucleotides together to form the new DNA
strand; often in two or three lines.
Many students failed to read the question carefully and did not answer the question as set. They
wrote at length about DNA replication, starting with DNA helicase. These answers were awarded a
maximum of two marks, because the question specifically asked how the complementary strand of
HIV DNA is made. Many students appeared to believe that DNA actively pulls free nucleotides into
place and makes them base pair; some even wrote about condensation reactions. There were
students who confused transcription with replication and gave accounts of mRNA production.
Some students appeared to focus on ‘HIV’ and ‘replication’ and gave an extended account of how
HIV infects cells, uses reverse transcriptase to make DNA, incorporates its DNA into host DNA,
takes over the cell, is replicated by the host cell, infects new cells and leads to AIDS. They often
went onto an additional page, or wrote their answer under 1.3 on the next page, in breach of
instructions given on the front of the exam pap
0 1 . 3 Contrast the structures of DNA and mRNA molecules to give three differences.
[3 marks]
1. DNA double stranded/double helix and mRNA single-stranded; 2. DNA (very) long and RNA short; 3. Thymine/T in DNA and uracil/U in RNA; 4. Deoxyribose in DNA and ribose in RNA; 5. DNA has base pairing and mRNA doesn’t/ DNA has hydrogen bonding and mRNA doesn’t; 6. DNA has introns/non-coding sequences and mRNA doesn’t;
In 1.3, it was pleasing to find that many students did obey the command word to ‘contrast’ and
gave full statements about the differences between DNA and RNA. Many students knew enough
about the structures of DNA and mRNA to give correct contrasting features and 47% obtained all
three marks.
0 7 . 1 Human papilloma virus (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.
[4 marks]
1. Vaccine/it contains antigen (from HPV); 2. Displayed on antigen-presenting cells; 3. Specific helper T cell (detects antigen and) stimulates specific B cell; 4. B cell divides/goes through mitosis/forms clone to give plasma cells; 5. B cell/plasma cell produces antibody;
7.1 was a very good discriminator. 19% obtained all four marks, 14% failed to score and equal
percentages obtained one, two or three marks. Many students had the idea that a vaccine contains
antigen and knowledge of antigen-presenting cells was common. There were also many correct
statements about plasma/B cells releasing antibodies. Fewer students had the idea of a B cell
dividing to form plasma cells. Not many students were able to express clearly the idea of a specific
helper T cell or B cell detecting, or responding to, a specific antigen. Quite a few students got
confused between the roles of T cells and B cells. Some students wrote at length about memory
cells and secondary responses, neither of which was required to answer the question. As in some
other questions, this inclusion of irrelevant material often generated additional pages and wasted
time that could have been spent on other questions.
Figure 5 shows a test that has been developed to find out if a person has antibodies to the human immunodeficiency virus (HIV) antigen. Figure 5 This test only detects the presence of HIV antibodies. Give two reasons why it cannot be used to find out if a person has AIDS. [2 marks] 0 7 . 1 HIV antigens are attached to a test well in a dish. A sample of blood plasma is added to the well. If HIV antibodies are present, they bind to the HIV antigen. The well is washed. A second antibody with an enzyme attached is then added. This binds specifically to the HIV antibody. The well is washed again. A yellow solution is added, which changes to blue if the enzyme is present. A blue colour shows that the person has HIV antibodies. Step 1 Step 2 Step 3 Step 4 7
This test only detects the presence of HIV antibodies. Give two reasons why it cannot
be used to find out if a person has AIDS.
[2 marks]
(To diagnose AIDS, need to look for/at)
- (AIDS-related) symptoms;
- Number of helper T cells;
he solution will remain yellow if a person is not infected with HIV. Explain why.
[2 marks]
- HIV antibody is not present;
- (So) second antibody/enzyme will not bind/is
not present;
A mother who was infected with HIV gave birth to a baby. The baby tested positive
using this test. This does not prove the baby is infected with HIV.
Explain why.
[2 marks]
- Children receive (HIV) antibodies from their
mothers/maternal antibodies; - (So) solution will always turn blue/will always
test positive (before 18 months)
A control well is set up every time this test is used. This is treated in exactly the same
way as the test wells, except that blood plasma is replaced by a salt solution.
Use information from Figure 5 to suggest two purposes of the control well.
[2 marks]
(Shows that)
1. Only the enzyme/nothing else is causing a
colour change;
2. Washing is effective/all unbound antibody is
washed away;
Q1.(a) Give two ways in which pathogens can cause disease. (2)
- (Releases) toxins;
- Kills cells / tissues.
- Accept any reference to cell / tissue damage
Ignore infecting / invading cells
Q2.Whooping cough is a disease that affects some infants. Doctors collected data relating to
whooping cough between 1965 and 1996.
They collected data for:
• the number of cases of whooping cough reported
• the percentage of infants vaccinated against whooping cough.
(a) Suggest two reasons why the percentage of infants vaccinated decreased between
1973 and 1975. (2)
Any two from:
- (Decrease linked to) few(er) cases of whooping cough;
- (Decrease linked to) risk of / fear of side effects;
- Insufficient vaccine available / too expensive to produce / distribute.
- Too expensive unqualified is insufficient for mark
(b) Between 1980 and 1990, there were three peaks in the number of reported cases of
whooping cough. After 1981, the number of cases of whooping cough in each peak
decreased.
Use the information from the graph to suggest why (2)
(b) 1. Vaccination rate increases;
2. Fewer people to spread the disease / whooping cough / more people
immune / fewer susceptible.
2. Neutral − greater herd effect
2. Allow description of immune
Q Reject ‘resistant’.
(c) The percentage of the population vaccinated does not need to be 100% to be
effective in preventing the spread of whooping cough.
Suggest why. (2)
- More people are immune / fewer people carry the pathogen;
If neither point 1 or 2 awarded
Herd immunity = 1 mark - So susceptible / unvaccinated people less likely to contact infected
people.
