Download A-level Human Biology Question paper Unit 5 - Inheritance

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General Certificate of Education
January 2009
Advanced Level Examination
BIOLOGY/HUMAN BIOLOGY (SPECIFICATION A)
Unit 5 Inheritance, Evolution and Ecosystems
Thursday 22 January 2009
BYA5
1.30 pm to 3.00 pm
For this paper you must have:
● a ruler with millimetre measurements.
You may use a calculator.
For Examiner’s Use
Question
Mark
Question
Mark
1
Time allowed: 1 hour 30 minutes
2
Instructions
● Use black ink or black ball-point pen.
● Fill in the boxes at the top of this page.
● Answer all questions.
● You must answer the questions in the spaces provided. Answers
written in margins or on blank pages will not be marked.
● If you need extra space use page 20 for your answers.
● Do all rough work in this book. Cross through any work you do not
want to be marked.
Information
● The maximum mark for this paper is 75.
● The marks for questions are shown in brackets.
● You will be marked on your ability to use good English, to organise
information clearly and to use accurate scientific terminology where
appropriate.
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Total (Column 1) →
Total (Column 2) →
TOTAL
Examiner’s Initials
(JAN09BYA501)
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BYA5
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Answer all questions in the spaces provided.
1
(a) Table 1 gives some features of three kingdoms of living organisms.
Complete the table by giving the name of each kingdom.
Table 1
Features
Kingdom
Multicellular, cells have a nucleus, no cell wall
Thread-like hyphae, cells have a nucleus, cell wall made of chitin
Single-celled, cells have no nucleus, cell wall made of murein
(3 marks)
1
(b) The horse, Equus caballus, and the donkey, Equus asinus, are two closely related
animals. Table 2 gives some details of their classification, in hierarchical order.
Table 2
Taxon
Horse
Donkey
Kingdom
Animalia
Animalia
Chordata
Chordata
Mammalia
Mammalia
Perissodactyla
Perissodactyla
Equidae
Equidae
Equus
Equus
caballus
asinus
Species
1
(b)
(i) Name the class and the family to which the horse and the donkey belong.
Class .........................................................................................................................
Family .......................................................................................................................
(1 mark)
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1
(b)
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(ii) The diploid chromosome number in the horse is 64. The diploid chromosome
number in the donkey is 62.
Crosses between a male donkey and a female horse result in offspring called
mules. Explain why scientists still regard the horse and the donkey as distinct
species.
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(3 marks)
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2 The diagram shows the chromosomes in a cell during one stage of meiosis. The positions of
the alleles of four genes are shown. The genotype of the cell is Dd Ee Qq Rr.
q
qr
r
E
D
d
QR
R
Q
2
E
De
d
e
(a) Name the stage of meiosis shown in the diagram. Explain your answer.
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(2 marks)
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2
(b) At the time shown in the diagram, this cell contains 0.52 picograms of DNA.
What mass of DNA will be present in a gamete formed from this cell?
................................................ picograms
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(c)
(1 mark)
(i) List the genotypes of each of the gametes that can be formed from the cell shown
in the diagram.
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(2 marks)
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(c)
(ii) Explain the advantage to the species of producing gametes with different
genotypes.
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(2 marks)
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3 The diagram shows part of the light-independent reactions of photosynthesis.
Ribulose
bisphosphate
(RuBP)
3
+
2 Molecules
of
glycerate
3-phosphate
(GP)
CO2
2 Molecules
of
triose
phosphate
(a) Two substances formed in the light-dependent reactions are necessary for the
conversion of GP into triose phosphate. Name these two substances.
1 .........................................................................................................................................
2 .........................................................................................................................................
(2 marks)
Scientists supplied carbon dioxide containing the radioactive isotope 14C to a culture of
photosynthesising single-celled algae. The algae were kept in the light for 200 seconds, then
in the dark for a further 1500 seconds, and were then returned to the light.
The graph shows the amounts of radioactivity found in RuBP and in GP at different times
throughout the investigation.
Light
Dark
Light
Radioactivity
GP
RuBP
0
(06)
100
200
300
400
500
1700 1800 1900 2000 2100 2200
Time / s
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3
(b) Describe the evidence in the graph which supports each of the following statements.
3
(b)
(i) The process by which RuBP is used to make GP is light-independent.
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(2 marks)
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(b)
(ii) RuBP is formed using GP only if substances formed in the light-dependent
reactions are present.
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(2 marks)
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(c) The amount of radioactive GP falls between 250 and 1700 seconds. Suggest why.
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(1 mark)
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4 The diagram shows aerobic respiration of glucose.
The numbers indicate how many molecules of each named substance are formed or used
when one molecule of glucose is respired.
Glucose
2 ATP
Stage C
Stage A
2 Reduced NAD
6 ATP
4 ATP
2 Pyruvate
2 CO2
2 Reduced NAD
2 Acetylcoenzyme A
Stage B
6 ATP
2 Coenzyme A
6 Reduced NAD
18 ATP
2 Reduced FAD
4 ATP
2 ATP
4 CO2
Hydrogen ions and electrons
6 Oxygen
(08)
6 Substance Y
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(a)
(i) Name substance Y. ...................................................................................................
4
(a)
(ii) Name stage B. ..........................................................................................................
4
(a) (iii) Name the organelle in which stages B and C occur. ...............................................
(3 marks)
4
(b) How many carbon atoms are there in one molecule of pyruvate? Use evidence from
the diagram to explain your answer.
Number .......................................
Evidence .............................................................................................................................
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(2 marks)
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(c) Aerobic respiration of glucose gives a respiratory quotient (RQ) of 1.0. Use
information from the diagram to explain how.
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(2 marks)
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(d)
(i) For each molecule of glucose, how many molecules of ATP are formed by
oxidative phosphorylation?
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(1 mark)
4
(d)
(ii) ATP is better than glucose as an immediate energy source for cell metabolism.
Give one reason why.
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(1 mark)
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5 Isle Royale is an island about 56 km long and 11 km wide in the middle of Lake Superior.
The moose is a type of deer. It is the largest herbivore on the island. The wolf is the only
carnivore on the island capable of killing a moose.
The graph shows the numbers of moose and of wolves on Isle Royale from 1959 to 2006.
3000
2500
2000
Number
1500
of moose
1000
500
0
60
50
40
Number
of wolves
30
20
10
0
1955 1960 1965 1970 1975 1980 1985 1990 1995 2000 2005 2010
Year
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5
(a)
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(i) Describe and explain the relationship between the number of moose and the
number of wolves from 1965 to 1976.
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(a)
(ii) The mark-release-recapture method would not be suitable for estimating the size
of the wolf population on Isle Royale between 1988 and 1992. Use information
from the graph to suggest why.
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(a) (iii) If a disease killed all the wolves on the island. What would you expect to
happen to the number of moose over the next 15 years?
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(1 mark)
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A and a are alleles of a single gene which controls one characteristic in the moose.
In 1995, a survey showed that 91% of the moose on the island had the phenotype caused by
the dominant allele A.
5
(b) Use the Hardy-Weinberg equation, and information from the graph, to calculate the
expected number of moose, in 1995, that were heterozygous for this characteristic.
Number of moose = ....................... (4 marks)
5
(c)
(i) If allele a gives a selective advantage, its frequency would be expected to rise in
subsequent generations. Explain why.
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(2 marks)
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(c)
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(ii) Give two causes of a change in allele frequency other than the effect of selection.
1 ................................................................................................................................
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2 ................................................................................................................................
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(2 marks)
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6 The border collie is a breed of dog. Most border collies have fur that is black and white.
Some have fur that is brown and white. The allele for black fur, B, is dominant to that for
brown fur, b.
Another gene can affect the expression of the gene for black or brown fur. The recessive
allele, n, of this gene causes black fur to become ‘blue’ (a dark grey) and brown fur to
become ‘lilac’. The presence of the allele N results in normal black or brown fur colour.
Fur colour in collies is not sex-linked.
6
(a) A breeder crossed two collies with black fur. The first litter of puppies were
3 black,
6
(a)
1 brown,
1 lilac.
(i) The parents were crossed again to produce a second litter. Complete the genetic
diagram to show offspring genotypes. Use information from the diagram to
explain how blue puppies could be produced in this cross.
Parental genotypes ..................... and .......................
Male
gametes
Female
gametes
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(4 marks)
6
(a)
(ii) What is the probability of obtaining a blue, female puppy in this litter? Use your
answer to part (i) to explain your answer.
Probability of a blue, female puppy = ..................................................... (3 marks)
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(b) The breeder had one blue female, called ‘Megan’. He wanted to know if Megan was
homozygous for fur colour. In an attempt to determine this, he crossed Megan with a
lilac male. The first litter contained 5 blue puppies. A second cross with the same
male produced a litter of 3 blue and 4 lilac puppies.
6
(b)
(i) Why did the breeder choose a lilac male for this cross rather than any other
colour?
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(2 marks)
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(b)
(ii) A χ2 test can be used to determine whether the observed results fit the expected
1 : 1 ratio. Give a suitable null hypothesis which can be checked using the χ2 test.
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(1 mark)
6
(b) (iii) Complete Table 3 to calculate the value of χ2 for the results.
Table 3
Characteristic
Observed
(O)
Blue fur
8
Lilac fur
4
Expected
(E)
(O – E)
χ2 =
(O – E)2
(O – E)2
E
(O – E)2
∑ E =
(2 marks)
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(b) (iv) Use Table 4 to determine whether the null hypothesis can be supported. Explain
how you arrive at your answer.
Table 4
Probability value
Degrees of
freedom
0.99
0.95
0.10
0.05
0.01
0.001
1
0.0002
0.0039
2.71
3.84
6.63
10.83
2
0.020
0.103
4.61
5.99
9.21
13.82
3
0.115
0.352
6.25
7.81
11.34
16.27
4
0.297
0.711
7.78
9.49
13.28
18.47
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(a) Trees are sometimes described as ‘carbon sinks’. Suggest what this means.
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(b) Large areas of rainforest are being cut down and the land used for growing crops such
as maize.
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(b)
(i) Removing the trees and replacing them with maize plants could lead to a
decrease in the diversity of animal species. Explain how.
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(b)
(ii) After the trees are cut down, the land may not be suitable for growing maize for
more than a few seasons. Suggest why.
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(c) An atom of nitrogen, present in nitrogen gas in the atmosphere, can become part of an
amino acid molecule in a maize plant. Describe how.
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