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A2 Human Biology Unit 2
Text questions
| P34/Q1 |
Because of the sampling bias: Is the probability of the quadrat landing at any point of the surveyed area the same? This could depend on:
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| P35/Q2 |
Mean = 3.7 Median = 3.8 Mode = 3.8 The data do not follow a normal
distribution because Mean (3.7) is different than
Median (3.8) and Mode (3.8). |
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| P36/Q3 |
Smaller standard deviation. |
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| P37/Q4 |
Yes. |
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| P38/Q5 |
AABB AABb AAbb AaBB AaBb Aabb aaBB aaBb aabb |
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| P41/Q6 |
All the genes for a certain characteristics (eg.
wing length) which are present in a given
population. |
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| P41/Q7 |
1 - 0.6 (L) = 0.4 (l) |
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| P42/Q8 |
Hardy Weinberg principle predicts that the
frequencies of the alleles of a particular gene in a
particular population (i.e. values of p and q) will
stay constant from generation to generation. |
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| P43/Q9 |
Homozugous recessive genotypes will always express
themselves in phenotypes (= q2) |
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| P44/Q10 |
See the lower graph/curve. |
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| P46/Q11 |
Directional selection. |
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| P47/Q12 |
Better camouflage. |
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| P48/Q13 |
Rate of respiration in organisms with sickle cell
allele would be depressed by a reduced oxygen -
carrying capacity of RBC. This could affect the
organisms' chances of survival. « Fight or flight »
situations often require high amounts of energy to
be released in a short period of time. This requires
a good supply of oxygen to the respiring tissues,
which would not be possible due to reduced oxygen -
carrying capacity of RBC. |
Assignment
| P50/Q1 | a) |
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| b) | Glutamate molecule has lost some atoms of H when
new bonds have been created. |
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| c) | The enzyme is responsible for the addition of a
carboxyl group. |
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| P50/Q2 | Phospholipids are present in the biological/cell
membranes. The ability of the residue to bind with
blood cells can speed up the process of the
formation of a blood clot. |
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| P50/Q3 | a) | Carboxylase (which controls the synthesis of
glutamate) only works in the presence of vit. K.
Warfarin (competitive inhibitor of vit.K) inhibits
the synthesis of glutamate. The R group of glutamate
residue binds to phospholipids and in this way is
speeding up the clot formation. If there is a lack
of glutamate this process of clot formation will
impaired. |
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| b) | Large doses of vit. K will change the ratio
between vit. K and warfarin molecules. As a result
more vit. K molecules will be to molecules of
carboxylase which control synthesis of glutamate. |
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| P51/Q4 | a) | WsWs - advantage: small amounts of vit. K
required in the diet. |
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| b) | WsWr - advantage: resistant to warfarin,
moderate amounts of vit. K required in the diet can
be obtained in natural conditions. |
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| P51/Q5 | a) |
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| b) |
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| P51/Q6 | Offspring WsWr from Parental crosses WsWr x WsWs at the beginning had the advantage of resistance to warfarin and moderate requirements for vit. K in the diet. Offspring WsWs was selected against due to its low resistance to warfarin. Frequency of the genotype WsWr increased. Later Parental crossed WsWr x WsWr produced 50% WsWr genotypes (selected for) and 50% genotypes selected against (WsWs, WrWr). No further increase in the frequency of genotype WsWr followed. | |||||||||||
| P51/Q7 | Genotypes with the dominant allele for warfarin
resistance are not being selected for, because
warfarin poisoning is extremely rare in the human
population. |
Examinations
| P52/Q1 | a) |
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| b) |
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| c) |
Genetic variation →
one genetic variant confers an advantage on the
possessor of a favourable allele of a gene
→ the possessor of the favourable allele will
have more offspring than the possessor of only the
less favourable alleles
→ favourable alleles will be passed onto some
of the offspring →
the frequency of this allele in the population will
increase (i.e. natural selection has occurred). |
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| P52/Q2 | a) |
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| P52/Q3 | a) |
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| b) |
Cob length: 24 - 26 cm |
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| c) |
(i) Polygenic inheritance: characteristics
controlled by many genes located on the same or on
different chromosomes. (ii) Homozygotes (produced
by pure breeding) both in variety A and variety B
had cobs of different length (→there
must be some other genes which modify phenotypes
resulting from homozygous genotypes.) Similar
observation can be made when analysing frequencies
of phenotypes, which result from heterozygous
genotypes of F1. |
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| d) |
(i) Both parental phenotypes demonstrate he
differences in the cob length. This variation can also be found in the offspring, so it is likely to be passed down in genes (inherited). (ii) In spite
of pure breeding variety A and variety B demonstrate
variation which may result |
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| e) |
(i) Gene is a part of DNA present in a chromosome,
which controls a certain characteristic/feature of a living organism (ii) An allele is a form of a
gene. A gene must have at least 2 different alleles, |
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| f) |
(i) ww (ii) Ww |
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| g) |
(i) 0.7 (W), 0.3 (w) (ii) Frequency Ww = 2pq = 2 x 0.7 x 0.3 = 0.42 %Ww = 42% |
