This document summarizes several pedigree analyses involving inheritance patterns of traits: 1. A pedigree is presented where a woman has a 100% chance of passing on a defective gene to her daughter based on her genotype. Calculations are shown for determining the probability that her grandson will have the defective gene. 2. Multiple pedigrees demonstrate autosomal recessive inheritance patterns based on affected and unaffected family members. X-linked inheritance is ruled out. 3. The mode of inheritance is determined to be autosomal recessive for a pedigree where an affected mother passes the trait to all her children. Mitochondrial inheritance and variable expression are discussed. 4. Probabilities are calculated for heterozygous parents and

1. PEDIGREE ANALYSIS
CSIR – UGC NET PROBLEMS SOLVED
Dr. Valli Syam

2. NET June 2019

3. NET June 2019
Let us take the allele for this gene as A and a
Genotype of II 4 is given aa – The chance of this woman
passing her defective gene to III-1 is 100%

4. NET June 2019
I-1 and I-2 are heterozygous (Aa) for this trait –
evident from II-1

5. NET June 2019
The chance of II-3 being heterozygous can be calculated
as follows
There are 4 possibilities when I-1 and I-2 are
heterozygous
1 AA : 1 Aa : 1 aA : 1 aa – 4
1 AA : 2 Aa : 1 aa – 4
Since this man is not affected 1 aa should be ruled out
1 AA : 2Aa – 3
So only 3 possibilities
He being carrier (Aa) for this trait is =
2
3
II-3 (Aa) x II-4 (aa)
III-1 receiving recessive allele from II-3 =
1
2
III-1 receiving recessive allele from II-4 = 1
The chance of III-1 getting this defective gene is
2
3
x
1
2
x 1=
2
6
=
1
3


6. NET June 2019
ANS 2

7. NET June 2019
ANS 2
Unaffected father – Affected Daughter
Rules out X – Linked Inheritance – Option A and B ruled
out

8. NET June 2019
Mothers (I-2, II-4, III-7) passing on the trait to all her children –
characteristic of Mitochondrial Inheritance – Option C is right
The answer must be 2 which has option C and D.
Variable expression can be due to heteroplasmy deals with the
presence of more than one type of organellar genome
(mitochondrial DNA or plastid DNA) within a cell or individual –
Option D also right
The answer is 2 which has Option C and D


9. NET December 2018

10. NET December 2018
4 and 5 are heterozygous – evident from 7 being
homozygous recessive (Expressing the phenotype)

11. NET December 2018
4 and 5 are heterozygous – evident from 7 being
homozygous recessive
Aa (4) x Aa (5)
AA Aa aA aa
1 AA 2 Aa 1 aa – 4
6 being homozygous recessive can be ruled out as he is
not having that trait
So only 3 possibilities are there
1 AA 2 Aa – 3
The probability of 6 being heterozygous (Aa) is
2
3


12. NET December 2017

13. NET December 2017
Normal parents giving rise to affected child – Family 1
RECESSIVE
So Option 2 and 4 can be ruled out

14. NET December 2017
Normal father giving rise to affected daughter – Family 2
and Affected mother giving rise to normal son – Family 3
X-linked recessive ruled out – ie., Option 1 can be ruled out
The answer is Option 3 – Autosomal Recessive is the right
answer


15. NET December 2016
The following pedigree shows the inheritance
pattern of a trait
From the following select the possible mode
of inheritance and the probability that the
daughter in generation III will show the trait
1. X-linked recessive, probability is 1/2
2. X-linked recessive, probability is 1/4
3. Autosomal recessive, probability is 1/2
4. Autosomal recessive, probability is 1/3

16. NET December 2016
Affected mother giving rise to unaffected son –
Not X-linked recessive – Option 1 and 2 ruled out
So Autosomal recessive
The following pedigree shows the inheritance
pattern of a trait
From the following select the possible mode
of inheritance and the probability that the
daughter in generation III will show the trait
1. X-linked recessive, probability is 1/2
2. X-linked recessive, probability is 1/4
3. Autosomal recessive, probability is 1/2
4. Autosomal recessive, probability is 1/3

17. NET December 2016
This woman being heterozygous is 100% - Would have
received the defective gene from affected father
The following pedigree shows the inheritance
pattern of a trait
From the following select the possible mode
of inheritance and the probability that the
daughter in generation III will show the trait
1. X-linked recessive, probability is 1/2
2. X-linked recessive, probability is 1/4
3. Autosomal recessive, probability is 1/2
4. Autosomal recessive, probability is 1/3

18. NET December 2016
This man is Homozygous Recessive – Since he is carrying
the trait
The following pedigree shows the inheritance
pattern of a trait
From the following select the possible mode
of inheritance and the probability that the
daughter in generation III will show the trait
1. X-linked recessive, probability is 1/2
2. X-linked recessive, probability is 1/4
3. Autosomal recessive, probability is 1/2
4. Autosomal recessive, probability is 1/3

19. NET December 2016
This daughter receiving the defective gene from father (aa) is
100%
This daughter receiving the defective gene from Mother (Aa) is
50% - ie. ½, Since mother is heterozygous (one normal gene
and another defective gene)
So Option 3 is the Right Answer

The following pedigree shows the inheritance
pattern of a trait
From the following select the possible mode
of inheritance and the probability that the
daughter in generation III will show the trait
1. X-linked recessive, probability is 1/2
2. X-linked recessive, probability is 1/4
3. Autosomal recessive, probability is 1/2
4. Autosomal recessive, probability is 1/3

20. NET December 2015
The following pedigree chart shows inheritance
of a given trait
The trait can be called
1. Autosomal dominant
2. Autosomal recessive
3. X-linked dominant
4. Sex limited

21. Affected parents giving rise to unaffected children
Dominant pedigree – Option 2 Ruled out
More towards Option 1 and 3
NET December 2015
The following pedigree chart shows inheritance
of a given trait
The trait can be called
1. Autosomal dominant
2. Autosomal recessive
3. X-linked dominant
4. Sex limited

22. Affected father giving rise to unaffected daughter and affected
son
Not X-linked – Option 3 Ruled out
NET December 2015
The following pedigree chart shows inheritance
of a given trait
The trait can be called
1. Autosomal dominant
2. Autosomal recessive
3. X-linked dominant
4. Sex limited

23. While all the male individuals are affected, this male is not
affected – So not sex limited – Option 4 ruled out
So the answer is Autosomal Dominant – Option 1 is the Right
Answer
NET December 2015
The following pedigree chart shows inheritance
of a given trait
The trait can be called
1. Autosomal dominant
2. Autosomal recessive
3. X-linked dominant
4. Sex limited


24. NET December 2014

25. I-1 is heterozygous for this phenotype – evident from II-2
NET December 2014

26. II-3 inheriting defective gene from I-2 is 100%
Probability of this woman being heterozygous is 1
NET December 2014

27. Details of Genotype of III-4 is not given, If his parents are
heterozygous
1 AA : 1Aa : 1aA : 1 aa - 4 possibilities
Since he is also not showing the phenotype homozygous recessive
condition is ruled out – Only 3 possibilities are there
1 AA : 2Aa – 3
II-4 being heterozygous is
2
3
(provided his parents are heterozygous
– this information is not available)
The probabilities of carriers in the population is given as
1
3
Therefore, Probability of III-4 being heterozygous is
2
3
𝑥
1
3
=
2
9
NET December 2014

28. 1 AA : 1Aa : 1aA : 1 aa
The probability of the child from heterozygous parents getting the
phenotype is
1
4
The probability of the child born to this parent (II-3 and II-4) showing
the phenotype = Probability of parents being heterozygous x
probability of getting the disorder with heterozygous parents
= 1 𝑥
2
9
𝑥
1
4
=
1
18
So the answer is Option 2 is the Right Answer
NET December 2014


29. NET December 2013
The following pedigree shows the
inheritance pattern of a rare recessive
disorder with complete penetrance
A child from marriage between individuals
II-1 & II-3 will show the disorder only if the
parents carry the recessive allele. What is
the probability that the child will show the
disorder?
1. 1/9, and the probability of the parents to
carry the recessive allele is 2/3
2. 1/4, and the probability of the parents to
carry the recessive allele is 3/4
3. 1/16, and the probability of the parents
to carry the recessive allele is 2/3
4. 1/64, and probability of the parents to
carry the recessive allele is 3/4

30. I-1,2 and 3,4 are heterozygous – evident from II-1 and II-4
NET December 2013
The following pedigree shows the
inheritance pattern of a rare recessive
disorder with complete penetrance
A child from marriage between individuals
II-1 & II-3 will show the disorder only if the
parents carry the recessive allele. What is
the probability that the child will show the
disorder?
1. 1/9, and the probability of the parents to
carry the recessive allele is 2/3
2. 1/4, and the probability of the parents to
carry the recessive allele is 3/4
3. 1/16, and the probability of the parents
to carry the recessive allele is 2/3
4. 1/64, and probability of the parents to
carry the recessive allele is 3/4

31. The chance of II-2 being heterozygous can be calculated as follows –
Same for II-3 also
1 AA : 1 Aa : 1 aA : 1 aa – 4
1 AA : 2 Aa : 1 aa – 4
Since II-2 and II-3 are not affected 1 aa should be ruled out
1 AA : 2Aa – 3
So only 3 possibilities
The chance of being carrier (Aa) for this trait is =
2
3
The chance of both II-2 and II-3 being heterozygous is
2
3
x
2
3
=
4
9
The probability of getting the disorder is
1 AA : 1Aa : 1aA : 1 aa
The probability of getting the disorder is
1
4
NET December 2013
The following pedigree shows the
inheritance pattern of a rare recessive
disorder with complete penetrance
A child from marriage between individuals
II-1 & II-3 will show the disorder only if the
parents carry the recessive allele. What is
the probability that the child will show the
disorder?
1. 1/9, and the probability of the parents to
carry the recessive allele is 2/3
2. 1/4, and the probability of the parents to
carry the recessive allele is 3/4
3. 1/16, and the probability of the parents
to carry the recessive allele is 2/3
4. 1/64, and probability of the parents to
carry the recessive allele is 3/4

32. The chance of both II-2 and II-3 being heterozygous is
2
3
x
2
3
=
4
9
The probability of getting the disorder with heterozygous parents is
1
4
The probability of the child born to this parent showing the disorder =
Probability of parents being heterozygous x probability of getting the
disorder with heterozygous parents
=
4
9
x
1
4
=
1
9
So the answer is Option 1 is the Right Answer
NET December 2013
The following pedigree shows the
inheritance pattern of a rare recessive
disorder with complete penetrance
A child from marriage between individuals
II-1 & II-3 will show the disorder only if the
parents carry the recessive allele. What is
the probability that the child will show the
disorder?
1. 1/9, and the probability of the parents to
carry the recessive allele is 2/3
2. 1/4, and the probability of the parents to
carry the recessive allele is 3/4
3. 1/16, and the probability of the parents
to carry the recessive allele is 2/3
4. 1/64, and probability of the parents to
carry the recessive allele is 3/4


33. NET December 2012
The following pedigree represents inheritance of a trait in an
extended family
What is the probable mode of inheritance and which
individuals conclusively demonstrate this mode of
inheritance
1. Autosomal recessive, III-2,3 and IV-1,2 conclusively
demonstrate the mode of inheritance
2. Autosomal recessive I-1,2 and II-2 conclusively
demonstrate the mode of inheritance
3. Autosomal dominant, III-2,3 and IV-1,2 conclusively
demonstrate the mode of inheritance
4. X-linked recessive, II-3,4 and 5 conclusively
demonstrate the mode of inheritance

34. Unaffected parents (III-2,3) giving rise to affected children
(IV-1,2) – Not a dominant pedigree – It is a
Recessive pedigree - Option 3 ruled out
NET December 2012
The following pedigree represents inheritance of a trait in an
extended family
What is the probable mode of inheritance and which
individuals conclusively demonstrate this mode of
inheritance
1. Autosomal recessive, III-2,3 and IV-1,2 conclusively
demonstrate the mode of inheritance
2. Autosomal recessive I-1,2 and II-2 conclusively
demonstrate the mode of inheritance
3. Autosomal dominant, III-2,3 and IV-1,2 conclusively
demonstrate the mode of inheritance
4. X-linked recessive, II-3,4 and 5 conclusively
demonstrate the mode of inheritance

35. I-1,2 and II-2 do not provide a conclusive evidence for
autosomal recessive condition – This can also be an
autosomal dominant pedigree or Y linked pedigree –
Option 2 ruled out
NET December 2012
The following pedigree represents inheritance of a trait in an
extended family
What is the probable mode of inheritance and which
individuals conclusively demonstrate this mode of
inheritance
1. Autosomal recessive, III-2,3 and IV-1,2 conclusively
demonstrate the mode of inheritance
2. Autosomal recessive I-1,2 and II-2 conclusively
demonstrate the mode of inheritance
3. Autosomal dominant, III-2,3 and IV-1,2 conclusively
demonstrate the mode of inheritance
4. X-linked recessive, II-3,4 and 5 conclusively
demonstrate the mode of inheritance

36. Affected father (I-1) giving rise to affected son (II-2)
Not X-linked – Option 4 Ruled out
So the answer is Option 1 is the Right Answer
NET December 2012

The following pedigree represents inheritance of a trait in an
extended family
What is the probable mode of inheritance and which
individuals conclusively demonstrate this mode of
inheritance
1. Autosomal recessive, III-2,3 and IV-1,2 conclusively
demonstrate the mode of inheritance
2. Autosomal recessive I-1,2 and II-2 conclusively
demonstrate the mode of inheritance
3. Autosomal dominant, III-2,3 and IV-1,2 conclusively
demonstrate the mode of inheritance
4. X-linked recessive, II-3,4 and 5 conclusively
demonstrate the mode of inheritance