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AS Human Biology Unit 13
Text questions
| P220/Q1 |
White blood cells |
|
| P221/Q2 |
Each phosphate group has a negative charge. |
|
| P222/Q3 |
DNA probe attaches to the complementary base
sequence of the DNA fragment (if this sequence is
present). |
|
| P223/Q4 | a) |
Pancreatic enzymes are released into the lumen of
the small intestine. They are not used up or
hydrolysed. Finally, they will find their way with
faeces out of the body. |
| b) |
Pancreas gradually loses its ability to produce
enzymes, hence lower levels of pancreatic enzymes
found in the faeces. |
|
| P224/Q5 |
No mixing of test sample with the Benedict's
solution required, no heating, reagents in solid
form. |
|
| P226/Q6 |
If the rate of urine production is increased, the
volume of blood will drop, hence lower blood
pressure. |
|
| P226/Q7 |
Antibiotics are specific toxins affecting the
growth/life processes of bacterial cells (but not
human cells), therefore they are inharmful to human
cells. |
|
| P227/Q8 |
Bacteriostatic antibiotics slow down the growth of
bacterial cells. Bactericidal antibiotics kill
bacterial cells. |
|
| P230/Q9 |
Second introduction of antigen
→ second
response. Memory B-cells (formed during clonal
expansion in the primary response), which have
remained in the lymphatic system, divide now rapidly
to produce plasma cells and the appropriate/specific
antibody. |
|
| P231/Q10 |
Monoclonal antibodies are immobilised on an inert
base and test solutions are passed over them. If
molecules of the target enzyme are present in the
test solution, they will combine with the
immobilised monoclonal antibodies. A second type of
antibody, which has an enzyme attached to its
molecules, is then added. It combines only with
those immobilised antibodies that are linked to the
target enzyme. Finally, a substrate is added which
is converted to a coloured product by the enzyme
attached to the second type of antibody. The amount
of colour tells us how much of the target enzyme was
present in the test solution. |
Assignment
| P234/Q1 | Draw and annotate fig 13.17 |
|
| P234/Q2 | a) | 50%; chromosomes belonging to one pair (so
called homologous chromosomes) separate during
meiosis 1 (the process is called random assortment).
The probability of one or the other chromosome from
the pair to migrate to a given daughter cell is the
same (50%), hence 50% of the offspring will have the
chromosome. |
| b) | Yes, because one of her homologue chromosomes
(i.e. a chromosome belonging to one pair) has a
piece of DNA with 4 repeats; (the other chromosome
with 3 repeats of DNA was inherited from the
father). |
|
| P235/Q3 | a) | Restriction enzymes are specific. They cut DNA
at a specific place, i.e. where specific sequence of
bases is present. |
| b) | The probe with the complementary base sequence
will take position at the complementary fragment of
a single stranded DNA with a particular repeated
sequence of non-coding DNA. The presence of the
probe with radioactive "label" molecule can be then
detected/confirmed by the use of a photographic
plate. |
|
| P235/Q4 | a) | Only bird 1 is a descendant of the parents (M,F),
whose DNA results are shown in fig. 13.18. It is
because the pattern of the DNA bands is similar to
the pattern seen in the mother's (M) and father's
(F) DNA. |
| b) | Positive results of genetic fingerprinting
indicate a high likelihood of genetic relationship
between the individuals concerned. However, it is
possible that a number of repeats of non-coding
fragments of DNA is due to chance/coincidence,
rather than inheritance. |
Examinations
| P236/Q1 | a) |
(i) The blood clot may break loose from its
attachment (:embolus). It will travel in the blood
until it is trapped in a small artery, which then
becomes completely blocked (:embolism). The blockage
in the coronary artery will cause death of an area
of the cardiac muscle. (ii) An increased concentration of the enzyme would
suggest a damage to the cells of the cardiac muscle.
This results in a leakage of the enzyme from the
damaged cells to the blood. |
| P236/Q2 | a) |
(i) Penicillin (produced by fungus Penicillium) is
acting specifically against bacteria. (ii) Viruses do not have cellular structure, they
are not living cells, hence they are not affected by
streptomycin. Viruses use host cells' enzymes, not
their own ones. Streptomycin does not affect host
cell's enzymes. |
| b) |
(i) For transcription to take place, the two DNA
polynucleotide chains must temporarily separate over
a certain length. If the separation is impossible,
the process of transcription will not take place,
hence no mRNA synthesised, and consequently the
process of translation will not happen. (ii) If polynucleotide chains of DNA can't separate,
then replication will not take place
→ mitosis
impossible →
tumour cells don't divide. |
|
| P237/Q3 | a) |
Glucose oxidase; |
| b) |
Enzyme A, blue dye, enzyme B; |
|
| c) |
(i) Enzymes denature at increased temperatures.
(ii) Enzyme A is specific to glucose, not to other
sugars. |
