Below is a pedigree of a family in which the son has albinism, an autosomal rece

Question

Below is a pedigree of a family in which the son has albinism, an autosomal recessive disease. This family also has a daughter, but it is known that this daughter does not have albinism. An RFLP analysis was done using a probe that identifies two morphs, one (2 kb & 7kb) and the other 9kb. The morph is linked to the albinism gene with 20 map units of recombination frequency. Using the information provided by the pedigree and RFLP, what is the probability that the daughter is a carrier for albinism? Assume that the son inherited nonrecombinant chromosomes from their parents. Show your work and circle your answer.

Explanation / Answer

As mentioned Albinism is an autosomal recessive disease. So one has to get both the recessive alleles to express this disease.

Daughter is not having albinism, so she must be carrying only one "a" allele.

So, we can say the daughter is a carrier for albinism and she is heterozygous as she carrying "Aa".

So, by seeing a punette square, we can the change for the daughter to be heterozygous is 0.4+0.1=0.5.

Now as the daughter is not suffering from albinism, we will try to rule out that particular option, by subtracting 0.4 from 1.
Sum of all probabilities= 1-0.4aa=0.6.

The probability that the daughter is a carrier for albinism=0.5/0.6x100=83%As mentioned Albinism is an autosomal recessive disease. So one has to get both the recessive alleles to express this disease.

Daughter is not having albinism, so she must be carrying only one "a" allele.

So, we can say the daughter is a carrier for albinism and she is heterozygous as she carrying "Aa".

So, by seeing a punette square, we can the change for the daughter to be heterozygous is 0.4+0.1=0.5.

Now as the daughter is not suffering from albinism, we will try to rule out that particular option, by subtracting 0.4 from 1.
Sum of all probabilities= 1-0.4aa=0.6.

The probability that the daughter is a carrier for albinism=0.5/0.6x100=83%

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