HKDSE Biology – a modern approach 4
Answers to Coursebook Exercises
22
Basic Genetics
Check your progress
22.1 What is genetics?
22.2 Mendel’ s investigations on monohybrid inheritance
A.
Fill in the blanks
1.
(a)
gene
(b)
chromosome
(c)
alleles
(d)
phenotype
(e)
genotype
(f)
heterozygous
(g)
Punnett square
(h)
test cross
(1 mark each)
(Total: 8 marks)
B.
Multiple choice
1.
C
2.
D
3.
A
4.
A
(1 mark each)
(Total: 4 marks)
22.3 Mendel’s investigations on dihybrid inheritance
Multiple choice
1.
D
2.
D
(1 mark each)
(Total: 2 marks)
22.4 Inheritance in humans
22.5 Pedigree
Multiple choice
1.
B
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HKDSE Biology – a modern approach 4
Answers to Coursebook Exercises
2.
C
3.
B
4.
C
(1 mark each)
(Total: 4 marks)
22.6 Variations
Multiple choice
1.
C
2.
D
3.
C
(1 mark each)
(Total: 3 marks)
Revision Exercise
A.
Multiple choice
1.
C
2.
D
3.
B
4.
D
5.
D
6.
C
7.
B
8.
D
9.
A
(1 mark each)
(Total: 9 marks)
B.
Short question
1.
(a)
environment
(b)
heredity
(c)
environment
(d)
heredity and environment
(1 mark each)
(Total: 4 marks)
C.
Structured questions
1.
(a)
black
© 2010 Aristo Educational Press Ltd.
(1 mark)
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HKDSE Biology – a modern approach 4
Answers to Coursebook Exercises
(b)
black male horse – Bb
(1 mark)
reddish-brown female horse – bb
(1 mark)
(c)
Parents:
Gametes:
Bb
(Black)
B
F1 generation:
Phenotypic ratio:
(d)
X
bb
(Reddish brown)
b
Bb
Black
1
b
bb
Reddish brown
:
1
(3 marks)
(1 mark)
The theoretical ratio is based on probability.
(1 mark)
In other words, the more offspring are produced, the closer the ratio will be to the
expected one.
(1 mark)
(Total: 9 marks)
2.
(a)
(1/2 mark)
No.
If the trait is sex-linked dominant, the male offspring between individual 5 (XaXa) and
individual 6 (XAY)
must receive a recessive allele
(1 mark)
Xa
from his mother and a Y chromosome from his
father, thus he must not have the trait.
(1 mark)
(1/2
It is not shown in the pedigree as individual 14 gets the trait.
(b)
mark)
Individual 11 does not have this trait and therefore he is homozygous.
(1 mark)
He has received one recessive allele from each of his parents (3 and 4).
(1 mark)
Individuals 3 and 4 both have a recessive allele.
(1 mark)
As they show the trait which is dominant, they must be heterozygous (Aa).
(2 marks)
(c)
3/
4
(1 mark)
(d)
Aa : aa = 1 : 1
(1 mark)
(e)
non-identical twins / fraternal twins
(1 mark)
(Total: 11 marks)
3.
(a)
Her daughter (individual 3) has blood group O who must be homozygous (ii) (1 mark)
and she must have received one allele i from each of her parents.
(1 mark)
Her son (individual 4) has blood group AB and should have received an allele IA from
one parent and an IB from the other one.
(1 mark)
Since individual 1 has blood group A, he would not have IB allele.
(1 mark)
Therefore, the genotype of individual 2 must be IBi.
(1 mark)
(b)
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HKDSE Biology – a modern approach 4
Answers to Coursebook Exercises
Individual 5
IAi
F1 generation:
Gametes:
F2 generation:
Genotypic ratio:
Individual 6
IA
i
IA IB
IA i
1
:
IBi
X
1
:
IB
i
IB i
ii
1
:
1
(4 marks)
The chance for individual 7 having blood group O is 25%.
(1 mark)
(Total: 10 marks)
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