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The ability of the human body to break down the red color in beets is controlled by an autosomal dominant allele. The inability is recessive, detected by red coloration of the urine (we will call this phenotype 'secretor'). If a nonsecretor woman with a secretor father has children with a nonsecretor man who in a previous relationship had a secretor daughter, what is the probability that their first child will be...

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Solution :

It is given that a woman is a nonsecretor but the father of that woman is a secretor. It means that the father's genotype is recessive, i.e. "ss".

The genotype of the nonsecretor mother would be Ss, who have received one recessive allele from her father.

Now the nonsecretor man have a secretor daughter from his previous marriage. That means that he is a carrier nd his genotype is Ss.

Now, we can cross between the heterozygous parents (Ss x Ss) will have offspring with the following genotypes :

1 SS -- Nonsecretor

2 Ss -- Nonsecretor

1 ss -- secretor

Thus the probability that their first child will be :

a). a secretor girl

   [tex]$=\frac{1}{4} \times \frac{1}{2}$[/tex]

  [tex]$=\frac{1}{8}$[/tex]

b). Non secretor girl

     [tex]$=\frac{3}{4} \times \frac{1}{2}$[/tex]

     [tex]$=\frac{3}{8}$[/tex]

c). a secretor boy

   [tex]$=\frac{1}{4} \times \frac{1}{2}$[/tex]

   [tex]$=\frac{1}{8}$[/tex]

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