Download public exam_basic genetics_R1

ECF Saint Too Canaan College
Biology public exam exercise
Basic Genetics
Name: ___________________
Class: _______ (
)
Date: _____________________
DSE(Practice Paper)_2012_IB_Q.5
1.
The pedigree below shows the inheritance of colour blindness in a family:
In humans, colour blindness is a sex-linked trait. Based on the above pedigree, deduce the
genotype of the mother (B) with respect to colour vision.
(Marks will not be awarded for genetic diagrams.)
(5 marks)
Bio_public exam_basic genetics
P.1
CE 2001_I_Q.3(a)
2.
The following table shows the information about some of the characteristic of 20 students in a
class:
(a) (i)
Draw a bar chart to show the distribution of students with different blood groups.
(3 marks)
(ii)
What type of variation does blood group show?
(1 mark)
(b) In humans, the ability to roll the tongue is determined by a pair of alleles. Students 1 and 2
are sisters. Both of their parents are tongue rollers. Deduce, with reasons, whether the allele
for tongue rolling is dominant or recessive.
(Marks will not be awarded to genetic diagrams.)
(5 marks)
(c) Students 19 and 20 are twin brothers. Are they identical or non-identical twins? Give a
reason for your answer.
(2 marks)
Bio_public exam_basic genetics
P.2
CE 2002_I_Q.3(c)
3.
An inherited abnormality in humans is the occurrence of unusually short fingers. The photographs
below show a hand with normal fingers and a hand with this abnormality:
The length of fingers is controlled by a pair of alleles. The allele for short fingers probably arose
from a mutation of the allele for normal fingers. The following pedigree shows the inheritance of
this abnormality in a family:
(a) (i)
What is meant by mutation?
(1 mark)
(ii)
Explain how a mutation in a person may lead to a change in the metabolic activities of
his cells.
(2 marks)
(b) Deduce which character, normal fingers or short fingers, is dominant.
Explain your deduction.
(Marks will not be awarded for genetic diagrams.)
(5 marks)
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3.
(c) Individual 6 is going to marry a female homozygous for normal fingers. Draw a genetic
diagram to show the result of the cross.
(Use F to stand for the dominant allele and f for the recessive allele.)
(3 marks)
CE 2003_I_Q.2(a)
4.
Individuals of a certain type of plant produce either purple or white flowers. The flower colour is
controlled by a pair of alleles. A gardener carried out two crosses with this type of plant and the
results are shown below:
(a) Based on cross 1, deduce the dominant flower colour. Explain your deduction.
(5 marks)
(b) Use symbols to show the possible genotypes of the parents in cross 2. Define the symbols
you use.
(3 marks)
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P.4
4.
(c) If the purple-flower parent in cross 2 was self-pollinated (i.e. the stigma receives pollen
grains from the same plant) and a large number of offspring was produced, predict the
phenotypes of the offspring and their ratio.
(3 marks)
CE 2004_I_Q.3(a)(i)(ii)
5.
The following pictures show two plants of the same species. Plant A has green leaves. Plant B is a
new form recently discovered by a scientist; it has variegated leaves.
The scientist performed an experiment by self-crossing plant A. A large number of offspring were
obtained were obtained and they all produced green leaves. He then repeated the same procedure
with plant B and all the offspring produced variegated leaves.
(a) Assuming that the colour pattern of the leaves is controlled by a pair of alleles, what
deductions can be made from the above results regarding the genotypes of plants A and B?
Explain how you arrive at your deductions.
(3 marks)
(b) In order to find out which colour pattern is dominant, the scientist performed another
experiment by crossing plant A with plant B. Explain how the results of this cross would
enable him to determine the dominant phenotype.
(3 marks)
Bio_public exam_basic genetics
P.5
CE 2007_I_Q.5
6.
The shape of the human little finger can be straight or bent. The photograph below shows a hand
with a bent little finger.
The inheritance of the shape of the little finger is controlled by a pair of alleles. The following
pedigree shows the inheritance of this trait in a family:
(a) After studying the pedigree, a student could not determine which little finger shape is
dominant. However, he drew the following conclusion.
‘Either individual 1 or individual 2 must be heterozygous.’
Do you agree with this conclusion? Explain your answer with reference to the role of
gametes in inheritance.
(3 marks)
(b) Provided that the allele for the bent little finger is dominant, deduce the possible genotype(s)
of individual 4.
(4 marks)
Bio_public exam_basic genetics
P.6
6.
(c) Individuals 5 and 6 are going to have another child. What is the probability of their second
child having straight little fingers? Illustrate your answer with a genetic diagram.
(5 marks)
CE 2008_I_Q.9(a)(iii)(2)
7.
A scientist performed a genetic experiment by crossing two different wheat plants. Describe the
procedures done in order to ensure cross-pollination, but not self pollination to occur.
(3 marks)
AL 2001_I_Q.3(a),(b),(c)(i)
8. (a) Contrast the genetic control of ABO blood group and that of red-green colour blindness in
humans.
(5 marks)
Bio_public exam_basic genetics
P.7
8.
(b) When a man, Tom, of blood group A marries a woman, May, of blood group O, deduce the
chances that a child of blood group A would be born to this couple. Use genetic diagram(s) to
show your deduction.
(7 marks)
(c) In a traffic accident, the victim suffered from severe blood loss and needed blood transfusion.
It was known that the victim was of blood group B. Explain whether Tom would be a poss
ible blood donor for the victim.
(2 marks)
CE 2004_I_Q.4(b)
9.
In a family, the father and the son are colour-blind, whereas the mother has normal vision. The
couple wants to have another child. What is the probability of their second child, whether a boy or
a girl, being colour-blind? Use a genetic diagram to illustrate your answer.
(5 marks)
Bio_public exam_basic genetics
P.8
DSE_2012_IB_Q.8
10. The photograph below shows the appearances of some kernels of a corn:
The purple colour is produced by a pigmented layer within the kernels. If the layer is not
pigmented, the yellow colour of the inner tissue becomes visible. Whether the kernel is smooth
or wrinkled is due to the type of food stored inside it. Smooth kernels (starchy corn) store starch
while wrinkled kernels (sweet corn) store soluble sugars. The surface of the sweet corns
becomes wrinkled when the corn dries up.
(a) With reference to osmosis, explain why the kernels of sweet corn become wrinkled when
they dry up but the kernels of starchy corn remain smooth.
(4 marks)
(b) The two traits of the kernels are controlled by genes located on different homologous
chromosomes. The following diagram shows the result of a cross between two pure-bred
corn plants, one with purple and smooth kernels and the other with yellow and wrinkled
kernels.
Bio_public exam_basic genetics
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(i)
Based on the results of the cross, deduce which phenotypes are dominant.
(ii)
The F1 generation produced was crossed with pure-bred corn plants with yellow and
wrinkled kernels, as shown below:
Explain the results of the cross using Mendel’s law of inheritance.
(3 marks)
(4 marks)
(Marks will not be awarded for genetic diagrams.)
(c) When Mendel proposed how traits are inherited, chromosomes had not been yet
discovered. In your opinion, how did Mendel come up with his hypothesis?
(3 marks)
Bio_public exam_basic genetics
P.10
DSE_2013_IB_Q.4
11. Red-green colour blindness is an X-linked recessive trait in humans. Peter is red-green colour
blind while his daughter, Mary, is normal.
(a) Deduce Mary’s genotype without using a genetic diagram.
(4 marks)
(b) Mary is an expectant mother. The photomicrograph below shows the karyotype of her
foetus:
(i)
From the photomicrograph, can we deduce whether this foetus will be red-green
colour blind or not? Explain your answer.
(2 marks)
(ii)
Is the foetus a boy or girl? Explain your answer with reference to the
photomicrograph.
(3 marks)
Bio_public exam_basic genetics
P.11
DSE_2015_IB_Q.4
12. Roger is found to be suitable for donating blood to recipients with blood types different from his
own. However, he cannot receive a blood transfusion from this parents. The blood types of his
father and mother are A and B respectively.
(a) What is Roger’s blood type?
(1 mark)
(b) Given that:
IA represents the allele for producing antigen A on the surface of red blood cells
IB represents the allele for producing antigen B on the surface of red blood cells
i represents the allele that does not lead to the production of any antigens on the surface of
red blood cells
(i)
Using the above symbols, state Roger’s genotype.
(1 mark)
(ii) Using the above symbols, state the genotype of his parents.
(2 marks)
Father: ________________________________ Mother: ______________________________
(c) Explain why Roger cannot receive blood transfusions from his parents.
(3 marks)
Bio_public exam_basic genetics
P.12
ECF Saint Too Canaan College
Biology public exam exercise – Basic Genetics answer
1.
Being a sex-link trait, the allele for colour blindness is located on the X-chromosome.
As the X chromosome of individual D must come from the mother (B),
the mother must have an X chromosome with an allele for colour blindness.
Being normal, the mother must bear the allele for normal colour vision on the other X
chromosome.
Hence, the genotype of the mother must be heterozygous / XBXb (where B is the allele for
normal colour vision & b is the allele for colour blindness).
1
1
1
1
1
(5)
2.
(a)
(i)
Title (T)
½
Correct labelling of axis (L)
½, ½
Correct plotting of the bars (P): 2 bars correct, ½ mark; 3 bars, 1 mark; all 4 1½
bars, 1½ mark
(ii)
(b)
(c)
discontinuous variation
1
The parents of students 1 and 2 are tongue-rollers, so each of them must carry at
least one allele for tongue rolling.
Student 2 is a non-roller, so she must have received at least one allele for non-tongue
1
rolling from either of her parents.
Thus at least one of the parents is heterozygous.
In a heterozygous condition, only the dominant allele is expressed.
Thus the allele for tongue rolling is dominant.
1
1
1
1
They are non-identical twins
because their blood groups are different, implying that they are genetically different
1
1
(11)
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P.13
3.
(a)
(b)
(i)
Mutation is a sudden change in the genetic materials.
1
(ii)
A mutation may lead to the synthesis of new protein / failure to synthesis a
certain protein.
Since proteins may be enzymes or may have other function in the cell,
absence / change of the original protein will result in a change in the
metabolic activities of cells.
1
1
Individual 3 / 6 is normal, so she / he must have received at least one allele for
normal fingers from either of her / his parents (individual 1 or 2).
Individual 1 and 2 have short fingers, so each of them must carry at least one allele
for short fingers.
1
1
Thus, at least one of the individual 1 and 2 is heterozygous.
In the heterozygous condition, only the dominant character is shown.
Thus, short finger is the dominant character.
1
1
1
(c)
1
1
1
(11)
4.
(a)
The offspring in cross 1 have white flowers, so they must have received at least one
allele for white flower from either of the parents.
Since both parents have purple flowers, each of them must carry at least one allele
for purple flower.
Thus at least one of the parents is heterozygous.
In the heterozygous condition, only the dominant character is shown.
Thus purple flower is the dominant character.
1
1
1
1
1
(b)
F represents the allele for purple flower; f represents the allele for white flower.
The possible genotypes of the purple-flower parent are FF or Ff
and that of the white-flower parent is ff.
1
1
1
(c)
If the genotype of purple-flower parent is FF, all offspring will produce purple
1
flowers.
If the genotype is Ff, purple-flower offspring and white-flower offspring will be
formed and they would be in ratio of 3:1
1
1
(11)
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5.
(a)
(b)
Plant A is homozygous for the green-leaf allele.
Plant B is homozygous for the variegated-leaf allele.
This is because all the offspring of each plant have the same phenotype as the parent.
1
1
Both plants A and B are homozygous, but of different phenotypes.
When they are crossed, all their offspring will be heterozygous.
In heterozygous condition, the phenotype shown by the offspring is the dominant
phenotype.
1
1
1
1
(6)
6.
(a)
Yes
To produce offspring with different phenotypes
there must be two different combinations of gametes.
Therefore, either one of the parents must be heterozygous, producing two types of
gametes carrying different alleles.
OR
The parents are of different phenotypes, hence, one of the parents must be
homozygous recessive.
To produce offspring with different phenotypes
the other parent must be heterozygous, producing two types of gametes carrying
different alleles.
(b)
Individual 1 possesses straight little fingers, she must be homozygous recessive
and pass an allele for straight little fingers to individual 4.
Individual 4 possesses bent little fingers, she must have at least one allele for bent
little fingers.
Hence, individual 4 is heterozygous.
1
1
1
(1)
(1)
(1)
1
1
1
1
Bio_public exam_basic genetics
P.15
6.
(c)
Define symbols (S)
Let B be the allele for bent little fingers
1
and b be the allele for straight little fingers
1
1
1
The probability for their child to have straight little finger is 0.5/50%
1
(12)
7.
During the experiment, use a plastic bag to wrap the treated flowers except during the
following treatments.
Remove the anthers from the flowrs.
Use a brush to transfer the pollen grains from other flowers to the stigma of the treated
flowers.
1
1
1
(3)
8.
(a)
1
1
3
(1)
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8.
(b)
Definition of symbols (1)
Let IA be the dominant allele for antigen A production
Let i be the recessive allele for no antigen production
1
Tom's genotype can be IA IA or IA i (½)
May's genotype is ii (½)
1
1
1
1
1
1
(c)
Tom cannot be a donor, because Tom's red blood cells have antigen A which
will lead to clumping when react with anti-A antibody in victim's blood.
1
1
(14)
9.
Definition of genetic symbols: (1)
Let b be the recessive allele for colour blindness
Let B be the dominant allele for normal vision
Father
parental genotypes
gamete types
Mother
x
b
XY
XBXb
Xb, Y
XB, Xb
(2)
4
genotypes of offspring
XBXb
XbXb
XBY
XbY
phenotypes of offspring
normal
colour-blind
normal
colour-blind
girls
boys
50%
50%
probability
Correct genetic diagram
layout (1)
(1)
(1)
L=1
(max. 5)
(5)
Bio_public exam_basic genetics
P.17
1
10. (a) 

soluble sugars lower the water potential of the kernels (1) while
starch, being insoluble in water, does not affect the water potential of the kernels
(1)

during development, kernels of sweet corns draw in a larger amount of / more
water by osmosis than those of starch kernels (1)

later on, when the kernels dry up, kernels of sweet corn shrinks as a result of
losing large amount of water (1)

as the skin is not elastic (1), thus the kernels of sweet corns become wrinkled
(any four of the above)
(b) (i)

as both parents are pure-bred plants, the F1 produced are all heterozygous
(1)

in heterozygous condition, dominant alleles will be expressed while
recessive alleles will be masked (1)
therefore, purple colour and smooth surface are the dominant phenotypes
(1)

(ii) 




as the individuals in F1 generation are heterozygous with and the genes for
the two traits are located on different chromosomes, the alleles for the
traits are assorted independently (1)
only one type of gametes for the pre-bred corn plant with yellow and
wrinkled kernels (1)
and four different types of gametes will be resulted from the F1 generation
(1)
after random fertilization, zygotes with four different combinations of
genotypes will be produced (1)
leading to the expressions of all the possible phenotypes (A, B, C and D)
in the F2 generation (1)
1
1
1
1
1
1
1
1
1
1
1
(c) 

Mendel collected a large amount of experimental evidence (1)
and based on his observation on the number of offspring with different
1
1

phenotypes (1)
he worked out a possible explanation for the observation by logical deduction
(1)
1
(14)
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P.18
11.
(a) -
as Mary’s father is red-green colour blind, he must have a X chromosome bearing
the recessive allele for red-green colour blindness (1)
-
being a female, Mary must have inherited the X chromosome bearing the
recessive allele from her father (1)
on the other hand, Mary is normal, therefore she must have a X chromosome
bearing dominant allele for normal eye sight (1)
therefore, Mary is a heterozygote (1)
(b) (i)
No, it is because
- the photomicrograph cannot reveal the genetic make-up of the foetus / the
photomicrograph only shows the number or appearance of chromosomes (1)
- whereas red-green colour blindness is due to the presence of a certain allele
(4)
(2)
(1)
(ii) -
the foetus is a female (1) because
the 23rd pair of chromosomes in the karyotype have similar size / length (1)
and thus the sex chromosomes of the foetus are both X chromosomes (1)
(3)
9
12.
(a) blood type O (1)
(1)
(b) (i) ii (1)
(ii) Father: IAi (1)
(1)
(2)
Mother: IBi (1)
(c) - antibodies A and B are already present in Roger’s blood (1)
(3)
- these antibodies will act against the antigens of the parents’ red blood cells (1)
- and causes blood cells clumping / hemolysis of blood cells (1) if Roger receive
blood transfusion from his parents.
7 marks
Bio_public exam_basic genetics
P.19