Question

2. 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

Let us assume that for the PTC tasting locus, the dominant allele is 'T' whose phenotype will be taster and recessive allele is 't' whose phenotype will be non-taster and for the albinism locus, the dominant allele be 'A' whose phenotype will be normal pigmentation and the recessive allele is 'a' whose phenotype will be albino.

Given in the question is a male with a heterozygous condition for tasters and albino i.e., his genotype will be TtAa marries a woman with albino and heterozygous for taster i.e., her genotype will be Ttaa. Assuming that these two loci assort independently then a cross between these two genotypes will produce the offsprings in the ratio of :

taster,pigmented:taster,albino:non-taster,pigmented:non-taster,albino=3:3:1:1

The cross of the two genotypes is given in the following table:

female\male	TA	Ta	tA	ta
Ta	TTAa	TTaa	TtAa	Ttaa
phenotype	taster,pigmented	taster, albino	taster,pigmented	taster,albino
ta	TtAa	Ttaa	ttAa	ttaa
phenotype	taster,pigmented	taster,albino	non-taster,pigmented	non-taster,albino

a) The probability of producing 3 tasters with normal pigmentation and 3 non-taster albino in a family of 6 children can be calculated in the following way:

The ratio of the taster,pigmented offspring is 3/8 and the ratio of non-taster albino is 1/8, which implies that the probability of producing 3 taster, pigmented and 3 non-taster,albino is 3 in 8 and 1 in 8.

Multiply all these probabilities and you will get the probability that is required i.e.

3/8*3/8*3/8*1/8*1/8*1/8=27/64

b) The probability of producing 3 taster, pigmented and 3 non-taster,albino in the same order is:

This is a combined probability i.e, the probabilities should be added and then the numerator should be multiplied by 1/4 as only four phenotypes are possible:

3/8+3/8+3/8+1/8+1/8+1/8=12/8

12/8*1/4=12/32=3/8

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