The following pedigree is of a family with an extremely rare genetic disorder. A

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Question

The following pedigree is of a family with an extremely rare genetic disorder. Assume that no new mutations have arisen in the pedigree and that the trait displays full penetrance.

a. What is the mode of inheritance of this trait (i.e., dominant versus recessive/X?linked versus autosomal)? Justify your answer.

b. Assign genotypes to every member in the pedigree using ?D? for the dominant allele and ?d? for the recessive.

c. If the two cousins IV?4 and IV?5 marry, what is the probability that their first child will express the trait. Show your calculation.

Explanation / Answer

1. The mode of inheritance of the trait for rare genetic disorder in the pedigree is dominant versus recessive.

If the mode of inheritance was X linked versus autosomal only the male child would be affected but not the female since defective X chromosome from carrier mother would be inherited by the son but the daughters will not be susceptible to the disease due to presence of another normal X chromosome. In this case even father does not carry the X linked allele since he is unaffected.

When the mode of inheritance of trait is dominant versus recessive both son and daughter have equal possibility of inheriting the trait for rare genetic disorder and the presence of one dominant defective allele might be sufficient to exhibit the disorder.

2. The above pedigree indicates that the trait for rare genetic disorder is autosomal recessive inheritance. Since both the parents in first generation are (I 1 and I2) unaffected but carriers since their children are affected and there are no affected individuals in third generation due to heterozygosity but might be carriers of the diseased allele which can affect individuals of next generation.

Considering the above facts genotypes assigned to every member in the pedigree using D for the dominant allele and d for the recessive are as follows:

Member

Possible genotype

I 1

I 2

II 1 (Affected)

II 2

II 3

II 4 (Affected)

II 5 (Affected)

II 6

DD or Dd

III 1

III 2

III 3

III 4

III 5

III 6

III 7

III 8

IV 1

IV 2 (Affected)

IV 3

IV 4

IV 5

IV 6

DD or Dd

3. Considering the cousins IV 4 and IV 5 as heterozygotic carriers of diseased allele the gametes produced and the possible genotypes of their children are represented in the Punnets square:

Dd x Dd

Female Gametes

Gametes

DD (Normal child)

Dd (Normal, Carrier child)

dd(Affected child)

Male Gametes Children

Hence from Punnets square probability of their first child being expressing the trait for genetic disorder is (1 out of 4).

Member

Possible genotype

I 1

I 2

II 1 (Affected)

II 2

II 3

II 4 (Affected)

II 5 (Affected)

II 6

DD or Dd

III 1

III 2

III 3

III 4

III 5

III 6

III 7

III 8

IV 1

IV 2 (Affected)

IV 3

IV 4

IV 5

IV 6

DD or Dd

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