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Sagot :
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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