Pedigree Questions from VCAA Exams (includes answers)
2009
The following pedigree is relevant for Questions 1 and 2.
Question 1
From an analysis of this pedigree, individuals with the same mitochondrial DNA (mDNA)
include
A. Kathy and Peter.
B. Peter and Jethro.
C. Jay, Rob and Jethro.
D. Jackson, Elliot and Dashiell.
58% of people got this correct
Question 2
Jackson is known to have the XH allele.
It is reasonable to claim that the allele must be also present in
A. Kate.
B. Peter.
C. Elliot.
D. Jethro.
84% of people got this correct
2008
The following information relates to Questions 13 and 14.
Phenylketonuria (PKU) is a condition in which affected individuals fail to produce the enzyme
phenylalanine hydroxylase. PKU is inherited as an autosomal recessive condition and the alleles
involved are
P : normal production of the enzyme
p : lack of enzyme
The following pedigree shows a family in which two members have PKU.
Question 13
In the pedigree shown, individuals that must be heterozygous for PKU include
A. I–2
B. I–4
C. II–1
D. II–6
85% of people got this correct
Question 14
The chance that individual III–3 is Pp with regard to the PKU locus is
A.1/3
B.1/2
C.2/3
D.3/4
21% of people got this correct
Question 21
Examine the following pedigree. Shaded individuals in the pedigree have a genetic disease.
Individuals marrying into the family, that is individuals II–1, II–4 and II–6, have no history of
the disease in their families.
The mode of inheritance of the trait shown in the pedigree is most likely to be
A. autosomal recessive.
B. autosomal dominant.
C. X-linked recessive.
D. X-linked dominant.
62% of people got this correct
Question 2
Some gene loci have several alternative alleles. With these gene loci, the DNA profi le of an
individual may be given in terms of the relative sizes of the alleles. For example, gene locus
THO has three alleles that are called THO 7, THO 9 and THO 10.
In a maternity hospital, a mother claimed that she had accidentally been given the wrong baby
boy. A DNA analysis of two gene loci, locus FGA and locus DS, was carried out on members of
the family and the results are included in the following pedigree. Assume that no mutation
involving these two loci has occurred in the family.
a. Does the data support the mother’s claim that she has been given the wrong baby boy?
Give reasons to support your answer.
Yes, The newborn baby boy is DS 11, 15. Either allele could have come from the mother.
Neither allele could have come from the father.
(2 marks)
60% of people got 0, 31% got 1, 9% got full marks
Many trials involving the technique of gene therapy have been carried out in humans, with
varying success.
b. What is the aim of gene therapy?
Gene therapy aims to correct gene function either by inserting a normal allele into a cell or by
inserting siRNA to prevent the translation of a faulty allele.
(1 mark)
34% of people got this correct
The technique of gene therapy also involves the use of vectors.
c. Name one agent that could act as a vector in gene therapy.
Virus or liposome.
(1 mark)
30% of people got this correct
In 1999, during a gene therapy trial for a particular enzyme deficiency, a young man died from
multiple organ failures four days after starting the treatment. He was in excellent health before
participating in the trial.
d. Suggest one event that could have occurred within the young man as a result of him receiving
gene therapy that resulted in his death.
Virus could have reverted to active state
Or
Insertion of gene disrupted another essential gene
(A suitably explained immune response, such as infection contracted during the procedure, was
also acceptable).
(1 mark)
29% of people got this correct
2007
Question 1
In the following pedigree the shaded individual has the trait under investigation.
The mode of inheritance of the trait is
A. autosomal dominant.
B. autosomal recessive.
C. X-linked dominant.
D. X-linked recessive.
77% of people got this correct
Question 4
The pedigree below represents a family of rabbits. The shaded rabbits have an inherited disease.
The phenotypes of rabbits I-1, I-2, I-3 and I-4 are not known.
On the basis of the offspring produced by II-1 and II-2 it has been suggested that the disease is
inherited as an X-linked dominant characteristic.
a. What evidence from generations II and III supports the suggestion made about the mode of
inheritance of the disease?
In generation II, the male is affected and the trait was passed to generation III where all females
but no males were affected.
(1 mark)
72% of people got this correct
b. If the mode of inheritance suggested is correct, complete the table below to show the possible
phenotypes and genotypes of rabbits I-1, I-2, I-3 and I-4. For each rabbit’s phenotype, select
from
 has the disease
 does not have the disease
 impossible to tell from the information given.
Use the following symbols to represent the alleles involved: XD, Xd, Y.
Impossible to tell
Does not have the disease
XDXd, XdXd
XdY
Has the disease
XDXd, XDXD
Impossible to tell
XDY, XdY
(4 marks)
19% of people 0, 22% got 1, 18% got 2, 19% got 3, 22% got 4
Assume each female in generation III was mated to a male from the same litter.
c. What genotypic and phenotypic ratios would you expect in the offspring?
Genotypic ratio: 1 XdXD: 1 XDY : 1 XdXd : 1 XdY
Phenotypic ratio: 1 affected female: 1 affected male: 1 normal female: 1 normal male
(2 marks)
51% of people 0, 26% got 1, 23% got 2
2006
Question 13
Achondroplasia is an autosomal dominant trait in humans that results in a form of dwarfism. In
some cases a child with achondroplasia is born to parents who have normal height.
The most likely reason for the appearance of the child with achondroplasia is that
A. the parents are carriers and the child has inherited the mutant allele from each parent.
B. a mutation has occurred in a gamete of either the mother or the father.
C. a mutation has occurred in a somatic cell of one of the parents.
D. a mutation has occurred in the tissues of the child.
53% of people got this correct
Question 3
Haemophilia is an X-linked recessive condition. The following pedigree shows a portion of a
family in which some members have haemophilia. Those on the pedigree with haemophilia are
shaded.
a. Use appropriate allele symbols from XH, Xh and Y to indicate the genotype of each of the
following individuals.
I1 __________XHXh___________________ and I2_________XhY_____________________
(1 mark)
b. The couple II1 and II2 have a son. What is the probability that the child has haemophilia?
Chance is 1 or 100% (that is, all sons will be haemophiliac)
(1 mark)
c. The couple II5 and II6 have a son. What is the probability that the child has haemophilia?
Chance is ½ or 50% of a child with haemophilia.
(1 mark)
For 3a-c 28% of people 0, 21% got 1, 16% got 2, 36% got 3
The following pedigree is of a family in which one member (shaded) has an autosomal recessive
condition. The alleles of the gene locus involved are G and g.
d. Give a possible genotype for each of the four members of the family.
I1
Gg
I2
Gg
II1
gg
II2
GG or Gg
(2 marks)
15% of people 0, 12% got 1, 73% got 2
2005
Question 3
The pedigree below represents the ancestry of a white female tiger, V.1, bred in a zoo. As with
domestic animals inbreeding is quite common. For example on this pedigree II.1 is mated to her
brother, II.2, and IV.2 mated with his sister, IV.3.
In tigers the combination of sex chromosomes is similar to that in humans.
a. With reference to the pedigree, explain why the locus for white coat colour cannot be
i. X-linked recessive
Cannot be X-linked recessive because father (II-2) of III-1 is unaffected.
ii. X-linked dominant
Cannot be X-linked dominant because an affected father (I-1) has an unaffected daughter.
(1+ 1= 2 marks)
53% of people 0, 25% got 1, 22% got 2
b. Explain whether there is enough information on the pedigree to distinguish between autosomal
dominant and autosomal recessive inheritance for white coat colour.
There is enough information to identify autosomal recessive, because individual III-1 is a white
tiger (affected) with parents with normal pigmentation (unaffected).
*Answers that stated the condition skipped a generation were not awarded marks, as specific
reference to the pedigree was required.
(1 mark)
36% of people got this correct
c. There is now a worldwide register of animals at risk of extinction. This is to avoid mating
between related animals. In a conservation program, explain why avoiding inbreeding is
important.
Inbreeding leads to a reduction in the number of phenotypes/reduction in diversity/reduction in
variation, and populations with less variation are at risk if selection pressures change.
(1 mark)
24% of people got this correct
2004
Use the following information to answer Questions 6 and 7.
Two linked genes in humans have the following alleles.
Gene 1 A : secretor a : non-secretor
Gene 2 B : production of protein X b : no protein X produced
Information regarding these traits in a family was collected and the chromosome composition of
each member for these two genes was established. It was found that, in all, there were four
different chromosome compositions in the family. The four types are shown in the following
table.
The pedigree for the family is shown below.
The code for interpreting the symbols is
Question 6
It is reasonable to conclude that the genotype of individual
A. I–1 is type 1.
B. II–2 is type 2.
C. III–2 is type 3.
D. III–12 is type 4.
Question 7
Assuming that all generation III individuals marry partners that are homozygous recessive for
these two genes, it is reasonable to conclude that individuals
A. 5 and 6 have an equal chance of having a non-secretor child.
B. 6 and 12 have an equal chance of having a non-secretor child.
C. 5 and 6 have an equal chance of having a child who fails to produce protein X.
D. 6 and 12 have an equal chance of having a child who fails to produce protein X.
73% of people got this correct
Question 2
Ptosis is a weakness in the muscles of the eyelid.
a. Using the information in the text box below, construct a pedigree to fit the description of the
inheritance of ptosis in the family.
Bill had ptosis. Bill married Jill who did not have ptosis and they had two children, a boy Ben who
was unaffected and a girl Daisy who had ptosis.
Daisy married Bob who did not have ptosis and they had three daughters, two girls with ptosis and
one girl without ptosis.
One mark was awarded for the pedigree and one mark for the key.
(2 marks)
6% of people go 0, 35% got 1, 59% got 2
b. The pedigree below shows another family in which ptosis was found.
i. Based on the information in this pedigree, what is the most likely mode of inheritance of
ptosis?
The most likely mode of inheritance of ptosis is autosomal dominant.
ii. Provide two pieces of evidence from the pedigree that support this conclusion.
Evidence 1
Two affected (I-1 and I-2) have unaffected children, so the trait must be dominant. If it was not,
I-1 and I-2 could not have unaffected children.
Evidence 2
There is an affected father (I-1) with unaffected daughters (II-2 and II-3), therefore the trait is
not X-linked dominant.
*To be awarded full marks, the answer needed to have a piece of evidence for dominance and a
piece of evidence for autosomal.
(1 + 2 = 3 marks)
36% of people got 0, 17% got 1, 25% got 2, 21% got 3.
2003
Use the following information to answer Questions 5–7.
The pedigree in Figure 2 shows the inheritance of Tay Sachs disease, a progressive neurological
defect in humans.
Question 5
It is reasonable to conclude from the pedigree that the trait shown is
A. X-linked recessive.
B. X-linked dominant.
C. autosomal recessive.
D. autosomal dominant.
78% of people got this correct
Question 6
The chance that a fourth child of III–1 and III–2 is male, and affected with Tay Sachs disease, is
A. 1/8
B. 1/4
C. 1/2
D. 3/4
40% of people got this correct
Question 7
The chance that IV–2 is heterozygous for Tay Sachs disease is
A. 1/4
B. 1/2
C. 2/3
D. 3/4
22% of people got this correct
2002
Use the following information to answer Questions 1 and 2.
Phenylketonuria (PKU) is inherited as an autosomal recessive condition. In the following
pedigree (Figure 1) shaded individuals have this trait.
Question 1
With respect to the PKU gene locus, you could reasonably conclude that individual
A. I – 1 is homozygous.
B. I – 2 is homozygous.
C. II – 5 is heterozygous.
D. III – 1 is heterozygous.
78% of people got this correct
Question 2
If II – 1 and II – 2 have another child, the chance that they will have a child affected by PKU is
A. 1/4
B.1/3
C.1/2
D. 1
81% of people got this correct
Question 2
Coat colour in cocker spaniel dogs varies. Four of these colours are black, liver, red and lemon.
These four colours result from the interaction of two particular autosomal genes.
The pedigree below (Figure 9) shows the inheritance of coat colour in a group of cocker spaniels.
a. I–1 and I–2 are heterozygous at both the R and B locus. What evidence from the pedigree
supports this conclusion?
Presence of red and liver in the offspring (or offspring includes dogs that are rrB- or R-bb). II-1
must be homozygous rr and II-3 must be homozygous bb.
*Students who did not refer directly to the pedigree and name individuals showing particular
traits were not awarded full marks.
(2 marks)
43% of people got 0, 41% got 1 and 16% got 2
b. i. What is the specific genotype of II–4?
RrBb
ii. What is the specific genotype of III–4?
rrBb
(1 + 1 = 2 marks)
30% of people got 0, 9% got 1, 61% got 2
c. Explain how many different phenotypes could be expected in the offspring of a mating
between individuals II–4 and III–4. Show all working.
RrBB or Bb = same phenotype (3
black)
Rrbb (1 red)
rrBB or Bb = same phenotype (3
liver)
rrbb (1 lemon)
The question specifically asked that students show all working. One mark was given for the
working and 1 mark for showing the connection between the genotype and phenotype
(2 marks)
50% of people got 0, 17% got 1, 33% got 2