Question

A man heterozygous at both the locus for PTC tasting and at the locus for albinism marries an albino woman heterozygous at the PTC tasting locus. These two loci assort independently. a) what is the probability of producing 3 tasters with normal pigmentation and 3 non-taster albinos in a family of 6 children? b) what is the probability of producing 3 tasters with normal pigmentation and 3 non- taster albinos in that order in a family of 6 children?

Answer 1

a). The probability of producing 3 tasters with normal pigmentation and 3 non-taster albinos in a family of 6 children = 0.002 = 0.2%

Explanation:

PTC = T; Non-PTC = t

Normal pigmentation = A; Albino = a

AaTt x aaTt ---Parents

	aT	at
AT	AaTT (normal pigmentation, PTC tasters)	AaTt (normal pigmentation, PTC tasters)
At	AaTt (normal pigmentation, PTC tasters)	Aatt (normal pigmentation, non-taster)
aT	aaTT (albino, taster)	aaTt (albino, taster)
At	aaTt (albino, taster)	aatt (albino, non-taster)

Normal pigmentation, PTC tasters = A_T_ = 3/8

Normal pigmentation, non-taster = A_tt = 1/8

Albino, taster = aaT_ = 3/8

Albino, non-taster = aatt = 1/8

The probability of producing 3 tasters with normal pigmentation and 3 non-taster albinos in a family of 6 children = 0.002 = 0.2%

= n!/s!t! p^s q^t

n = 6                                

s = 3 (taster, normal pigmentation)

t = 3 (non-taster, albino)

p = 3/8 (taster, normal pigmentation probability)

q = 1/8 (non-taster, albino probability)

= 6!/3!3! (3/8)^3 (1/8)^3

= 720/36 (27/512) (1/512)

= 20 * 0.0527 * 0.00195

= 0.002

b). The probability of producing 3 tasters with normal pigmentation and 3 non- taster albinos in that order in a family of 6 children = 0.0001 = 0.01%

= (3/8)^3 (1/8)^3

= 0.0527 * 0.00195

= 0.0001