Question

모모모 < 모 2 1 | 미모 모 오 or lo III = - -

Answer 1

PART 1

| | | 白白

a) the disorder is most likely to be an autosomal recessive disorder.

For the disorder to be autosomal dominant either of the parents should had been affected (carriers not possible for dominant traits).

If the disorder was X-linked recessive, the mother in generation I would have been a carrier and would have produced all diseased male progenies, or alternatively, the father in generation I would have been diseased and would have passed the allele to all his daughters.

If the disorder was X-linked dominant either of the parents should have been affected.

b) Genotypes (AA = normal, Aa = carrier, aa = diseased)

I1-Aa, I2-Aa

II1-AA, II2-Aa, II3-aa, II4-AA/Aa, II5-aa, II6-AA/Aa, II7-Aa, II8-AA

III1-AA/Aa, III2-AA/Aa, III3-Aa, III4-Aa, III5-AA/Aa

IV1-AA/Aa, IV2-AA/Aa, IV3-Aa, IV4-Aa, IV5-AA/Aa

c) Genotype of individual 1 and 2 should be Aa.

therefore, all possible genotypes that can be observed in their offspring are AA, Aa, Aa, aa

Probability of a child being healthy is $\frac{3}{4}$ [AA or Aa]

Probability of a child being diseased is $\frac{1}{4}$ [aa]

P(only 1 affected child out of 5 children) = 100 3 3 3 1 81 *7*7** 7 * 1024

d) the males of generation IV are not diseased which means their genotype is either AA or Aa. Genotype of an affected female would be aa.

P(an af fected child) = P (of genotype Aa in male of generation IV) P(of genotype aa in child) = 5x = |-

PART 2

??? 2 口口口 - = = 2

a) The most probable inheritance pattern is of X-linked dominant trait as an affected father is producing all affected daughters.

b) For the given pedigree chart.

all the affected males have genotype $X^{D}Y$

all the affected females have genotype

XDX

all the unaffected females have genotype

all the unaffected males have genotype

c) genotype of individual 1 is . genotype of individual 2 is $X^{D}Y$

[only the daughters will be affected as they will get the disease allele from their father and the allele being dominant will manifest]

P(individual 1 and 2 will have an affected child) =

d) in any case probability of getting a son is 50% i.e. 1/2

here, out of all four possible progenies (, $X^{D}Y$ ,
XDX
, ) only 2 are sons

therefore, out of 2 sons there is only one possibility of getting a diseased son

:: Pof a disea sed son ) =